From a4500beaca85f81bccc5e0c349287c5d3cf35c92 Mon Sep 17 00:00:00 2001 From: Sri Hari Date: Fri, 7 Mar 2025 00:19:01 +0530 Subject: [PATCH 1/5] Batch-6/Neetcode-150/Added-hints --- hints/burst-balloons.md | 39 ++++++++++++++++++ hints/buy-and-sell-crypto-with-cooldown.md | 39 ++++++++++++++++++ hints/coin-change-ii.md | 39 ++++++++++++++++++ hints/count-paths.md | 39 ++++++++++++++++++ hints/count-subsequences.md | 39 ++++++++++++++++++ hints/edit-distance.md | 39 ++++++++++++++++++ hints/gas-station.md | 31 ++++++++++++++ hints/hand-of-straights.md | 31 ++++++++++++++ hints/interleaving-string.md | 39 ++++++++++++++++++ hints/jump-game-ii.md | 39 ++++++++++++++++++ hints/jump-game.md | 39 ++++++++++++++++++ hints/longest-common-subsequence.md | 31 ++++++++++++++ hints/longest-increasing-path-in-matrix.md | 23 +++++++++++ hints/longest-increasing-subsequence.md | 47 ++++++++++++++++++++++ hints/maximum-product-subarray.md | 31 ++++++++++++++ hints/maximum-subarray.md | 39 ++++++++++++++++++ hints/merge-triplets-to-form-target.md | 31 ++++++++++++++ hints/partition-equal-subset-sum.md | 39 ++++++++++++++++++ hints/partition-labels.md | 39 ++++++++++++++++++ hints/regular-expression-matching.md | 39 ++++++++++++++++++ hints/target-sum.md | 31 ++++++++++++++ hints/word-break.md | 31 ++++++++++++++ 22 files changed, 794 insertions(+) create mode 100644 hints/burst-balloons.md create mode 100644 hints/buy-and-sell-crypto-with-cooldown.md create mode 100644 hints/coin-change-ii.md create mode 100644 hints/count-paths.md create mode 100644 hints/count-subsequences.md create mode 100644 hints/edit-distance.md create mode 100644 hints/gas-station.md create mode 100644 hints/hand-of-straights.md create mode 100644 hints/interleaving-string.md create mode 100644 hints/jump-game-ii.md create mode 100644 hints/jump-game.md create mode 100644 hints/longest-common-subsequence.md create mode 100644 hints/longest-increasing-path-in-matrix.md create mode 100644 hints/longest-increasing-subsequence.md create mode 100644 hints/maximum-product-subarray.md create mode 100644 hints/maximum-subarray.md create mode 100644 hints/merge-triplets-to-form-target.md create mode 100644 hints/partition-equal-subset-sum.md create mode 100644 hints/partition-labels.md create mode 100644 hints/regular-expression-matching.md create mode 100644 hints/target-sum.md create mode 100644 hints/word-break.md diff --git a/hints/burst-balloons.md b/hints/burst-balloons.md new file mode 100644 index 000000000..c553a4f8e --- /dev/null +++ b/hints/burst-balloons.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n ^ 3) time and O(n ^ 2) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ Try to simulate the process recursively by passing the array to the recursive function. At each step, iterate through the array, pop an element, and recursively apply the same process to the two subarrays on both sides of the popped element, returning the maximum result from all recursive paths. This approach is exponential. Can you think of a way to optimize it? Maybe you should consider observing the subproblems instead of modifying the array. +

+
+ +
+
+ Hint 2 +

+ Instead of passing the array, we can pass the range of indices l and r that need to be processed. We pad the input array with 1s on both sides for easier computation, but l and r represent the first and last indices of the original input array. Can you think of a reverse engineering approach for popping elements? +

+
+ +
+
+ Hint 3 +

+ We determine the result by considering each element as the last one to be popped in the current range. For each element, we calculate its value by multiplying it with the elements at l - 1 and r + 1, then recursively solve the subproblems for the ranges (l, i - 1) and (i + 1, r), where i is the current element in the given range. Can you think of a way to optimize and avoid redundant calculations? +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid redundant calculations. A hash map or a 2D array can be used to store results since the recursive function parameters l and r are within the range of the input array size. +

+
\ No newline at end of file diff --git a/hints/buy-and-sell-crypto-with-cooldown.md b/hints/buy-and-sell-crypto-with-cooldown.md new file mode 100644 index 000000000..84988bd54 --- /dev/null +++ b/hints/buy-and-sell-crypto-with-cooldown.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree. Can you determine the possible decisions at each recursion step? Also, can you identify the base cases and the essential information that needs to be tracked during recursion? +

+
+ +
+
+ Hint 2 +

+ At each recursion step, we can buy only if we haven't already bought a coin, or we can sell if we own one. When buying, we subtract the coin value, and when selling, we add it. We explore all possible buying and selling options recursively, iterating through the coins from left to right using index i. For the cooldown condition, if we buy a coin, we increment the index i by two. +

+
+ +
+
+ Hint 3 +

+ We can use a boolean variable canBuy to indicate whether buying is allowed at the current recursive step. If we go out of bounds, we return 0. This approach is exponential. Can you think of a way to optimize it? +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid recalculations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/coin-change-ii.md b/hints/coin-change-ii.md new file mode 100644 index 000000000..3f18a6318 --- /dev/null +++ b/hints/coin-change-ii.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n * a) time and O(n * a) space, where n is the number of coins and a is the given amount. +

+
+ +
+
+ Hint 1 +

+ As we need to find the total number of combinations, think in terms of recursion and visualize it as a decision tree where multiple coin choices are available at each recursion step. Can you determine a way to allow picking the same coin multiple times? Maybe you should consider the decisions made at each recursion step. +

+
+ +
+
+ Hint 2 +

+ The given coins are unique. We recursively iterate through the coins array using index i, tracking the collected amount along the current path. At each step, we can either skip the current coin or pick it, ensuring the total does not exceed the target. To allow picking the same coin multiple times, we recurse with the same index but an updated amount, generating different combinations. +

+
+ +
+
+ Hint 3 +

+ If we reach the target amount, we return 1. The recursion stops if the index goes out of bounds. We count all possible ways and return the total. This approach is exponential. Can you think of a way to optimize it? Maybe you should consider an approach to avoid redundant computations. +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid redundant computations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/count-paths.md b/hints/count-paths.md new file mode 100644 index 000000000..299793547 --- /dev/null +++ b/hints/count-paths.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m is the number of rows and n is the number of columns in the grid. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree, where we have two choices at each step. Can you determine the base condition and recurrence relation? +

+
+ +
+
+ Hint 2 +

+ We recursively traverse the grid using row i and column j. At each step, we explore both possibilities: moving down or moving right, ensuring we do not go out of bounds. If we reach the bottom-right cell, we return 1. +

+
+ +
+
+ Hint 3 +

+ This approach has exponential complexity. Can you think of a way to optimize the recursion? Maybe you should consider using a dynamic programming approach. +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid recalculations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/count-subsequences.md b/hints/count-subsequences.md new file mode 100644 index 000000000..53a37dcee --- /dev/null +++ b/hints/count-subsequences.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m is the length of the string s and n is the length of the string t. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree, as we need to explore all subsequences of s. Can you determine the possible decisions at each recursion step? +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the strings using indices i and j for s and t, respectively. At each recursion step, we can either skip the current character of s or include it in the current path if it matches the current character of t. Can you determine the base conditions for this recursive function? +

+
+ +
+
+ Hint 3 +

+ If index j goes out of bounds, we return 1 as a valid subsequence is found. If index i goes out of bounds first, we return 0. At each recursion step, we return the sum of both paths. This approach is exponential. Can you think of a way to optimize it? +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid redundant computations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/edit-distance.md b/hints/edit-distance.md new file mode 100644 index 000000000..277fa3cbe --- /dev/null +++ b/hints/edit-distance.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m and n are the lengths of the strings word1 and word2, respectively. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree, as we have choices at each recursion step. Can you determine the recurrence relation and base cases? +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the strings using indices i and j for word1 and word2, respectively. If the characters at the current indices match, we increment both indices without counting an operation. Otherwise, we have three choices: insert the character at the current index of word1 (increment j), delete the current character of word1 (increment i), or replace the character at index i in word1 (increment both i and j). +

+
+ +
+
+ Hint 3 +

+ If index i goes out of bounds, we return the number of remaining characters in word2 (using insert operations). If index j goes out of bounds, we return the number of remaining characters in word1 (using delete operations). At each step, we return the minimum operation path. This approach is exponential. Can you think of a way to optimize it? +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results and avoid redundant calculations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/gas-station.md b/hints/gas-station.md new file mode 100644 index 000000000..52464072d --- /dev/null +++ b/hints/gas-station.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to start at each gas station, simulate the process, and return the index where completing the circuit is possible. This would be an O(n^2) time solution. Can you think of a better way? Maybe a greedy approach works. +

+
+ +
+
+ Hint 2 +

+ We can immediately return -1 if sum(gas) < sum(cost), as completing the circuit is impossible due to insufficient gas. Otherwise, a solution always exists because the total gas is sufficient to cover the total cost, meaning there must be a valid starting point that allows completing the circuit. +

+
+ +
+
+ Hint 3 +

+ We start with a variable total to track the gas balance and initialize the result index to 0. As we iterate through the array with index i, we accumulate the difference (gas[i] - cost[i]). If total becomes negative at any index, we reset it to 0 and update the result index to (i + 1). +

+
\ No newline at end of file diff --git a/hints/hand-of-straights.md b/hints/hand-of-straights.md new file mode 100644 index 000000000..1a35e0287 --- /dev/null +++ b/hints/hand-of-straights.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(nlogn) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ It is observed that to form a group, the minimum value should be the starting value of the group. Additionally, the minimum value in the array serves as the starting value for one or more groups based on its frequency. Can you think of an efficient way to determine the frequencies of array elements? Maybe a specific data structure can be useful here. +

+
+ +
+
+ Hint 2 +

+ We can use a hash map to store the elements along with their frequencies. Additionally, we sort the given array. Then, we iterate through the sorted array and try to form groups by decrementing the frequency count. If we fail to form a group at any step, we immediately return false. Can you think why this works? +

+
+ +
+
+ Hint 3 +

+ Sorting ensures we start with the smallest available value, while the hash map helps track element availability using frequency counts. At each step, we pick the smallest available value x and attempt to form a group from x to x + groupSize - 1. If all elements are present based on their frequency counts, we decrement their counts as we iterate. If we successfully form all groups, we return true; otherwise, we return false. +

+
\ No newline at end of file diff --git a/hints/interleaving-string.md b/hints/interleaving-string.md new file mode 100644 index 000000000..16a8c100a --- /dev/null +++ b/hints/interleaving-string.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m is the length of the string s1 and n is the length of the string s2. +

+
+ +
+
+ Hint 1 +

+ If the sum of the characters in s1 and s2 does not equal s3, we return false. Think in terms of recursion and visualize it as a decision tree, where we explore different combinations of portions from both strings. Can you determine the possible decisions at each recursion step? +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the strings using indices i, j, and k for s1, s2, and s3, respectively. At each step, we extend the current path in two directions based on whether the k-th character of s3 matches the current character of s1 or s2. If any path returns true, we immediately return true. If k goes out of bounds, it means we have successfully formed the interleaved string, so we return true. +

+
+ +
+
+ Hint 3 +

+ This approach is exponential. Can you think of a way to optimize it? Since k depends on i and j, it can be treated as a constant, as we can derive k using i + j. +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid redundant computations. Treating i and j as states, we can use a hash map or a 2D array to store the results. +

+
\ No newline at end of file diff --git a/hints/jump-game-ii.md b/hints/jump-game-ii.md new file mode 100644 index 000000000..4c7c41e30 --- /dev/null +++ b/hints/jump-game-ii.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to recursively explore all paths from index 0 to its reachable indices, then process those indices similarly and return the minimum steps to reach the last index. This would be an exponential approach. Can you think of a better way? Maybe a greedy approach works. +

+
+ +
+
+ Hint 2 +

+ We maintain two pointers, l and r, initially set to 0, representing the range of reachable indices. At each step, we iterate through the indices in the range l to r and determine the farthest index that can be reached from the current range. +

+
+ +
+
+ Hint 3 +

+ We then update the pointers l and r to the next range, setting l to r + 1 and r to the farthest index reachable from the current range. We continue this process until the pointers go out of bounds. +

+
+ +
+
+ Hint 4 +

+ The number of steps taken represents the minimum steps required to reach the last index, as it is guaranteed that we can reach it. +

+
\ No newline at end of file diff --git a/hints/jump-game.md b/hints/jump-game.md new file mode 100644 index 000000000..91e7092c9 --- /dev/null +++ b/hints/jump-game.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to recursively explore all paths from index 0 to its reachable indices, then process those indices similarly, returning true if we reach the last index. This would be an exponential approach. Can you think of a better way? Maybe a greedy approach works. +

+
+ +
+
+ Hint 2 +

+ Instead of processing the array from index 0, start from the last index. Let the target index be goal = n - 1. Iterate in reverse from index n - 2. +

+
+ +
+
+ Hint 3 +

+ At each iteration, we check whether the current index can reach goal. If it can, we update goal to the current index, as reaching the current index means we can also reach the goal. +

+
+ +
+
+ Hint 4 +

+ To determine if we can reach the last index, the goal should be 0 after the iteration. Otherwise, reaching the last index is not possible. +

+
\ No newline at end of file diff --git a/hints/longest-common-subsequence.md b/hints/longest-common-subsequence.md new file mode 100644 index 000000000..d25df25e5 --- /dev/null +++ b/hints/longest-common-subsequence.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m is the length of the string text1 and n is the length of the string text2. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree. Can you determine the possible decisions at each step? Maybe you should consider iterating through both strings recursively at the same time. +

+
+ +
+
+ Hint 2 +

+ At each recursion step, we have two choices: if the characters at the current indices of both strings match, we move both indices forward and extend the subsequence. Otherwise, we explore two paths by advancing one index at a time and recursively finding the longest subsequence. We return the maximum length between these two choices. This approach is exponential. Can you think of a way to optimize it? +

+
+ +
+
+ Hint 3 +

+ We return 0 if either index goes out of bounds. To optimize, we can use memoization to cache recursive call results and avoid redundant calculations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/longest-increasing-path-in-matrix.md b/hints/longest-increasing-path-in-matrix.md new file mode 100644 index 000000000..8008fe73f --- /dev/null +++ b/hints/longest-increasing-path-in-matrix.md @@ -0,0 +1,23 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(m * n) time and O(m * n) space, where m is the number of rows and n is the number of columns in the given matrix. +

+
+ +
+
+ Hint 1 +

+ If we move from one cell to an adjacent cell with a greater value, we won't revisit a cell, as the path is strictly increasing. A brute force approach would be to run a dfs from every cell and return the longest path found. This approach is exponential. Can you think of a way to optimize it to avoid redundant calculations? +

+
+ +
+
+ Hint 2 +

+ We can use memoization to cache the results of recursive calls and avoid redundant computations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/longest-increasing-subsequence.md b/hints/longest-increasing-subsequence.md new file mode 100644 index 000000000..36a4e3e78 --- /dev/null +++ b/hints/longest-increasing-subsequence.md @@ -0,0 +1,47 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n ^ 2) time and O(n ^ 2) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A subsequence is formed by selecting elements while maintaining their order. Using recursion, we can generate all subsequences. The recursive function returns the length of the longest increasing subsequence up to index i, processing from left to right. At each step, we decide whether to include or exclude the current element. +

+
+ +
+
+ Hint 2 +

+ Since the sequence must be increasing, we represent choices by adding 1 when including an element and 0 when excluding it. In recursion, how can we ensure the current element is greater than the previous one? Perhaps additional information is needed to process it. +

+
+ +
+
+ Hint 3 +

+ We can store the index of the previously chosen element as j, making it easier to process the current element at index i. If and only if j == -1 or nums[i] > nums[j], we include the current element and extend the recursive path. Can you determine the recurrence relation? At most, two recursive calls are made at each recursion step. +

+
+ +
+
+ Hint 4 +

+ We stop the recursion when index i goes out of bounds and return 0 since no more elements can be added. The initial recursion call starts with j = -1. At each step, we include the current element if it is greater than the previous one and continue the recursion, or we exclude it and explore the next possibility. We return the maximum value obtained from both paths. +

+
+ +
+
+ Hint 5 +

+ The time complexity of this approach is exponential. We can use memoization to store results of recursive calls and avoid recalculations. A hash map or a 2D array can be used to cache these results. +

+
\ No newline at end of file diff --git a/hints/maximum-product-subarray.md b/hints/maximum-product-subarray.md new file mode 100644 index 000000000..d46997d32 --- /dev/null +++ b/hints/maximum-product-subarray.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force solution would be to find the product for every subarray and then return the maximum among all the products. This would be an O(n ^ 2) approach. Can you think of a better way? Maybe you should think of a dynamic programming approach. +

+
+ +
+
+ Hint 2 +

+ Try to identify a pattern by finding the maximum product for an array with two elements and determining what values are needed when increasing the array size to three. Perhaps you only need two values when introducing a new element. +

+
+ +
+
+ Hint 3 +

+ We maintain both the minimum and maximum product values and update them when introducing a new element by considering three cases: starting a new subarray, multiplying with the previous max product, or multiplying with the previous min product. The max product is updated to the maximum of these three, while the min product is updated to the minimum. We also track a global max product for the result. This approach is known as Kadane's algorithm. +

+
\ No newline at end of file diff --git a/hints/maximum-subarray.md b/hints/maximum-subarray.md new file mode 100644 index 000000000..f8071b628 --- /dev/null +++ b/hints/maximum-subarray.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to compute the sum of every subarray and return the maximum among them. This would be an O(n^2) approach. Can you think of a better way? Maybe you should consider a dynamic programming-based approach. +

+
+ +
+
+ Hint 2 +

+ Instead of calculating the sum for every subarray, try maintaining a running sum. Maybe you should consider whether extending the previous sum or starting fresh with the current element gives a better result. Can you think of a way to track this efficiently? +

+
+ +
+
+ Hint 3 +

+ We use a variable curSum to track the sum of the elements. At each index, we have two choices: either add the current element to curSum or start a new subarray by resetting curSum to the current element. Maybe you should track the maximum sum at each step and update the global maximum accordingly. +

+
+ +
+
+ Hint 4 +

+ This algorithm is known as Kadane's algorithm. +

+
\ No newline at end of file diff --git a/hints/merge-triplets-to-form-target.md b/hints/merge-triplets-to-form-target.md new file mode 100644 index 000000000..fd475aa48 --- /dev/null +++ b/hints/merge-triplets-to-form-target.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ An important observation is that we can ignore triplets with values greater than the target triplet. +

+
+ +
+
+ Hint 2 +

+ Specifically, if a triplet t has any element greater than the corresponding value in target (i.e., t[0] > target[0], t[1] > target[1], or t[2] > target[2]), we can discard it. This is because using such a triplet in operations would exceed the target values, making it invalid. +

+
+ +
+
+ Hint 3 +

+ Now, from the remaining valid triplets, we only need to check whether the target triplet values exist. Since all values in the valid triplets are less than or equal to the corresponding values in the target triplet, finding the target triplet among them guarantees that we can achieve it. +

+
\ No newline at end of file diff --git a/hints/partition-equal-subset-sum.md b/hints/partition-equal-subset-sum.md new file mode 100644 index 000000000..6dfc98881 --- /dev/null +++ b/hints/partition-equal-subset-sum.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n * t) time and O(n * t) space, where n is the size of the input array and t is half the sum of the array elements. +

+
+ +
+
+ Hint 1 +

+ If the sum of the array elements is not even, we can immediately return false. Think in terms of recursion, where we try to build a subset with a sum equal to half of the total sum. If we find such a subset, the remaining elements will automatically form another subset with the same sum. The entire array can also be considered as one subset, with the other being empty. Can you visualize this as a decision tree to process the array recursively? +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the array with index i. At each step, we decide whether to include the current element in the subset or not. Instead of forming the subset explicitly, can you think of a better approach? Maybe you only need to track the subset sum rather than generating the subset itself. +

+
+ +
+
+ Hint 3 +

+ We can track the subset sum using a variable curSum. At each step, we make two recursive calls. If adding the current element does not exceed the target, we include it. If either path leads to a solution, we immediately return true. Can you determine the base case for this recursion? All elements in the array are positive. +

+
+ +
+
+ Hint 4 +

+ If curSum equals half the sum of the array elements, we return true. If index i goes out of bounds, we return false. This solution is exponential, but we can use memoization to cache recursive call results and avoid redundant computations. We can use a hash map or a 2D array with dimensions n * t, where n is the size of the input array and t is half the sum of the input array elements. +

+
\ No newline at end of file diff --git a/hints/partition-labels.md b/hints/partition-labels.md new file mode 100644 index 000000000..34cb90372 --- /dev/null +++ b/hints/partition-labels.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ +

+
+ +
+
+ Hint 1 +

+ +

+
+ +
+
+ Hint 2 +

+ +

+
+ +
+
+ Hint 3 +

+ +

+
+ +
+
+ Hint 4 +

+ +

+
\ No newline at end of file diff --git a/hints/regular-expression-matching.md b/hints/regular-expression-matching.md new file mode 100644 index 000000000..e17c78964 --- /dev/null +++ b/hints/regular-expression-matching.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(m * n) time and O(m * n) space, where m is the length of the string s and n is the length of the string p. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree, where we explore different combinations to match the strings when encountering *. Multiple decisions are made at each step to find a valid matching path. Can you determine the possible decisions at each recursion step? +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the strings using indices i and j for s and p, respectively. If the characters match or p[j] is '.', we increment both i and j to process the remaining strings. When the next character of string p is '*', we have two choices: skip it (treating it as zero occurrences) or match one or more characters (if s[i] matches p[j]), incrementing i accordingly. +

+
+ +
+
+ Hint 3 +

+ If both indices go out of bounds, we return true; otherwise, we return false. If any recursive path returns true, we immediately return true. This approach is exponential. Can you think of a way to optimize it? +

+
+ +
+
+ Hint 4 +

+ We can use memoization to cache the results of recursive calls and avoid redundant calculations. A hash map or a 2D array can be used to store these results. +

+
\ No newline at end of file diff --git a/hints/target-sum.md b/hints/target-sum.md new file mode 100644 index 000000000..06e9e9a3e --- /dev/null +++ b/hints/target-sum.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n * m) time and O(n * m) space, where n is the size of the input array and m is the sum of all the elements in the array. +

+
+ +
+
+ Hint 1 +

+ Try to think in terms of recursion and visualize it as a decision tree, where we have two choices at each recursion step: assigning a positive or negative sign. +

+
+ +
+
+ Hint 2 +

+ We recursively iterate through the array using index i, tracking the current sum along the recursive path. Each step branches into two paths, and we sum the number of ways to reach the target. If the index i goes out of bounds, we return 1 if the current sum equals the target; otherwise, we return 0. +

+
+ +
+
+ Hint 3 +

+ This approach is exponential. We can use memoization to cache recursive call results and avoid redundant calculations. A hash map or a 2D array with modifications can be used for caching. If using a 2D array, the dimensions can be (n * (2m + 1)), where n is the array size and m represents the sum of the array elements. +

+
\ No newline at end of file diff --git a/hints/word-break.md b/hints/word-break.md new file mode 100644 index 000000000..cb850a49c --- /dev/null +++ b/hints/word-break.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n * m * t) time and O(n) space, where n is the length of the string s, m is the number of words in wordDict, and t is the maximum length of any word in wordDict. +

+
+ +
+
+ Hint 1 +

+ Try to think of this problem in terms of recursion, where we explore all possibilities. We iterate through the given string s, attempting to pick a word from wordDict that matches a portion of s, and then recursively continue processing the remaining string. Can you determine the recurrence relation and base condition? +

+
+ +
+
+ Hint 2 +

+ The base condition is to return true if we reach the end of the string s. At each recursive call with index i iterating through s, we check all words in wordDict and recursively process the remaining string by incrementing i by the length of the matched word. If any recursive path returns true, we immediately return true. However, this solution is exponential. Can you think of an optimization? Maybe you should consider an approach that avoids repeated work. +

+
+ +
+
+ Hint 3 +

+ We can avoid recalculating results for recursive calls by using memoization. Since we iterate with index i, we can use a hash map or an array of the same length as s to cache the results of recursive calls and prevent redundant computations. +

+
\ No newline at end of file From 188f5d6394d10b1588dd030e013df58cda7348c8 Mon Sep 17 00:00:00 2001 From: Sri Hari Date: Fri, 7 Mar 2025 22:58:28 +0530 Subject: [PATCH 2/5] Batch-6/Neetcode-150/Added-hints --- articles/insert-new-interval.md | 12 +++++-- articles/merge-intervals.md | 4 ++- hints/insert-new-interval.md | 31 ++++++++++++++++++ hints/meeting-schedule-ii.md | 39 +++++++++++++++++++++++ hints/meeting-schedule.md | 31 ++++++++++++++++++ hints/merge-intervals.md | 39 +++++++++++++++++++++++ hints/minimum-interval-including-query.md | 31 ++++++++++++++++++ hints/non-overlapping-intervals.md | 39 +++++++++++++++++++++++ hints/partition-labels.md | 16 +++------- hints/valid-parenthesis-string.md | 39 +++++++++++++++++++++++ 10 files changed, 265 insertions(+), 16 deletions(-) create mode 100644 hints/insert-new-interval.md create mode 100644 hints/meeting-schedule-ii.md create mode 100644 hints/meeting-schedule.md create mode 100644 hints/merge-intervals.md create mode 100644 hints/minimum-interval-including-query.md create mode 100644 hints/non-overlapping-intervals.md create mode 100644 hints/valid-parenthesis-string.md diff --git a/articles/insert-new-interval.md b/articles/insert-new-interval.md index e35159e6e..367a09391 100644 --- a/articles/insert-new-interval.md +++ b/articles/insert-new-interval.md @@ -221,7 +221,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output list. --- @@ -515,7 +517,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output list. --- @@ -708,4 +712,6 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ \ No newline at end of file +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output list. \ No newline at end of file diff --git a/articles/merge-intervals.md b/articles/merge-intervals.md index 56f8a6313..4ff97b569 100644 --- a/articles/merge-intervals.md +++ b/articles/merge-intervals.md @@ -169,7 +169,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n \log n)$ -* Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm. +* Space complexity: + * $O(1)$ or $O(n)$ space depending on the sorting algorithm. + * $O(n)$ for the output list. --- diff --git a/hints/insert-new-interval.md b/hints/insert-new-interval.md new file mode 100644 index 000000000..a807b4c61 --- /dev/null +++ b/hints/insert-new-interval.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) extra space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ The given intervals are non-overlapping and sorted in ascending order based on the start value. Try to visualize them as line segments and consider how a new interval can be inserted. Maybe you should analyze different cases of inserting a new interval. +

+
+ +
+
+ Hint 2 +

+ First, we append all intervals to the output list that have an end value smaller than the start value of the new interval. Then, we encounter one of three cases: we have appended all intervals, we reach an interval whose start value is greater than the new interval’s end, or we find an overlapping interval. Can you think of a way to handle these cases efficiently? +

+
+ +
+
+ Hint 3 +

+ We iterate through the remaining intervals, updating the new interval if its end value is greater than or equal to the current interval's start value. We adjust the start and end of the new interval to the minimum and maximum values, respectively. After this, any remaining intervals are appended to the output list, and we return the result. +

+
\ No newline at end of file diff --git a/hints/meeting-schedule-ii.md b/hints/meeting-schedule-ii.md new file mode 100644 index 000000000..1342e260f --- /dev/null +++ b/hints/meeting-schedule-ii.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(nlogn) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ Try to visualize the meetings as line segments on a number line representing start and end times. The number of rooms required is the maximum number of overlapping meetings at any point on the number line. Can you think of a way to determine this efficiently? +

+
+ +
+
+ Hint 2 +

+ We create two arrays, start and end, containing the start and end times of all meetings, respectively. After sorting both arrays, we use a two-pointer based approach. How do you implement this? +

+
+ +
+
+ Hint 3 +

+ We use two pointers, s and e, for the start and end arrays, respectively. We also maintain a variable count to track the current number of active meetings. At each iteration, we increment s while the start time is less than the current end time and increase count, as these meetings must begin before the earliest ongoing meeting ends. +

+
+ +
+
+ Hint 4 +

+ Then, we increment e and decrement count as a meeting has ended. At each step, we update the result with the maximum value of active meetings stored in count. +

+
\ No newline at end of file diff --git a/hints/meeting-schedule.md b/hints/meeting-schedule.md new file mode 100644 index 000000000..0216b7b66 --- /dev/null +++ b/hints/meeting-schedule.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(nlogn) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ If two intervals are sorted in ascending order by their start values, they overlap if the start value of the second interval is less than the end value of the first interval. And these are called overlapping intervals. +

+
+ +
+
+ Hint 2 +

+ A brute force approach would be to check every pair of intervals and return false if any overlap is found. This would be an O(n^2) solution. Can you think of a better way? Maybe you should visualize the given intervals as line segments. +

+
+ +
+
+ Hint 3 +

+ We should sort the given intervals based on their start values, as this makes it easier to check for overlaps by comparing adjacent intervals. We then iterate through the intervals from left to right and return false if any adjacent intervals overlap. +

+
\ No newline at end of file diff --git a/hints/merge-intervals.md b/hints/merge-intervals.md new file mode 100644 index 000000000..89af36fe5 --- /dev/null +++ b/hints/merge-intervals.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(nlogn) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ Sorting the given intervals in ascending order based on their start values is beneficial, as it helps in identifying overlapping intervals efficiently. How can you determine if two intervals overlap? +

+
+ +
+
+ Hint 2 +

+ If two intervals are sorted in ascending order by their start values, they overlap if the start value of the second interval is less than or equal to the end value of the first interval. +

+
+ +
+
+ Hint 3 +

+ We iterate through the sorted intervals from left to right, starting with the first interval in the output list. From the second interval onward, we compare each interval with the last appended interval. Can you determine the possible cases for this comparison? +

+
+ +
+
+ Hint 4 +

+ The two cases are: if the current interval overlaps with the last appended interval, we update its end value to the maximum of both intervals' end values and continue. Otherwise, we append the current interval and proceed. +

+
\ No newline at end of file diff --git a/hints/minimum-interval-including-query.md b/hints/minimum-interval-including-query.md new file mode 100644 index 000000000..9b56faba2 --- /dev/null +++ b/hints/minimum-interval-including-query.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(nlogn + mlogm) time and O(n + m) space, where m is the size of the array queries and n is the size of the array intervals. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to iterate through each query and, for each query, check all intervals to find the result. This would be an O(m * n) solution. Can you think of a better way? Maybe processing the queries in sorted order could help. +

+
+ +
+
+ Hint 2 +

+ We sort the intervals by start value and process the queries in ascending order. Using a pointer i, we add intervals to a min-heap while their start values are less than or equal to the query, storing their end values and sizes. +

+
+ +
+
+ Hint 3 +

+ The min-heap is ordered by interval size. We remove elements from the heap while the top element’s end value is less than the current query. The result for the query is the top element’s size if the heap is non-empty; otherwise, it is -1. +

+
\ No newline at end of file diff --git a/hints/non-overlapping-intervals.md b/hints/non-overlapping-intervals.md new file mode 100644 index 000000000..1e07a9c0b --- /dev/null +++ b/hints/non-overlapping-intervals.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(nlogn) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ If two intervals are sorted in ascending order by their start values, they overlap if the start value of the second interval is less than the end value of the first interval. And these are called overlapping intervals. +

+
+ +
+
+ Hint 2 +

+ A brute force approach would be to sort the given intervals in ascending order based on their start values and recursively explore all possibilities. This would be an exponential approach. Can you think of a better way? Maybe a greedy approach works here. +

+
+ +
+
+ Hint 3 +

+ We first sort the given intervals based on their start values to efficiently check for overlaps by comparing adjacent intervals. We then iterate through the sorted intervals from left to right, keeping track of the previous interval’s end value as prevEnd, initially set to the end value of the first interval. +

+
+ +
+
+ Hint 4 +

+ We then iterate from the second interval. If the current interval doesn't overlap, we update prevEnd to the current interval's end and continue. Otherwise, we set prevEnd to the minimum of prevEnd and the current interval’s end, greedily removing the interval that ends last to retain as many intervals as possible. +

+
\ No newline at end of file diff --git a/hints/partition-labels.md b/hints/partition-labels.md index 34cb90372..c5e3551bb 100644 --- a/hints/partition-labels.md +++ b/hints/partition-labels.md @@ -2,7 +2,7 @@
Recommended Time & Space Complexity

- + You should aim for a solution with O(n) time and O(m) space, where n is the length of the string s and m is the number of unique characters in the string s.

@@ -10,7 +10,7 @@
Hint 1

- + A character has a first and last index in the given string. Can you think of a greedy approach to solve this problem? Maybe you should try iterating over one of these two indices.

@@ -18,7 +18,7 @@
Hint 2

- + We store the last index of each character in a hash map or an array. As we iterate through the string, treating each index as a potential start of a partition, we track the end of the partition using the maximum last index of the characters seen so far in the current partition. Additionally, we maintain the size of the current partition using a variable, say size.

@@ -26,14 +26,6 @@
Hint 3

- -

-
- -
-
- Hint 4 -

- + We update the end of the current partition based on the maximum last index of the characters, extending the partition as needed. When the current index reaches the partition’s end, we finalize the partition, append its size to the output list, reset the size to 0, and continue the same process for the remaining string.

\ No newline at end of file diff --git a/hints/valid-parenthesis-string.md b/hints/valid-parenthesis-string.md new file mode 100644 index 000000000..29baed7d5 --- /dev/null +++ b/hints/valid-parenthesis-string.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(n) time and O(n) space, where n is the length of the input string. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would try all possibilities when encountering a '*' and recursively solve the problem, leading to an exponential approach. Can you think of a better way? Maybe a data structure commonly used in parenthesis problems could be useful. +

+
+ +
+
+ Hint 2 +

+ We can solve the problem using a stack-based approach. We maintain two stacks: one for tracking the indices of left parentheses and another for star indices. As we iterate through the string from left to right, we push indices onto their respective stacks when encountering a left parenthesis '(' or a star '*'. Can you determine the logic for the right parentesis case? +

+
+ +
+
+ Hint 3 +

+ If the left parenthesis stack is not empty, we pop from it. Otherwise, we pop from the star stack, treating the star as a left parenthesis to keep the string valid. After iterating the string, the stacks might be non-empty? Can you determine the logic for this case? +

+
+ +
+
+ Hint 4 +

+ Now, we try to match the remaining left parentheses with stars, ensuring the stars appear after the left parentheses in the string. We simultaneously pop from both stacks, and if the index of a left parenthesis is greater than that of a star, the string is invalid as there is no matching right parenthesis. In this case, we return false. +

+
\ No newline at end of file From fc340c1dde4643133d4db8b09d2a04014e95c009 Mon Sep 17 00:00:00 2001 From: Sri Hari Date: Sat, 8 Mar 2025 22:20:12 +0530 Subject: [PATCH 3/5] Batch-6/Neetcode-150/Added-hints --- articles/plus-one.md | 14 +++++++----- articles/set-zeroes-in-matrix.md | 2 +- articles/spiral-matrix.md | 12 +++++++--- hints/non-cyclical-number.md | 23 +++++++++++++++++++ hints/plus-one.md | 23 +++++++++++++++++++ hints/rotate-matrix.md | 31 +++++++++++++++++++++++++ hints/set-zeroes-in-matrix.md | 39 ++++++++++++++++++++++++++++++++ hints/spiral-matrix.md | 31 +++++++++++++++++++++++++ 8 files changed, 165 insertions(+), 10 deletions(-) create mode 100644 hints/non-cyclical-number.md create mode 100644 hints/plus-one.md create mode 100644 hints/rotate-matrix.md create mode 100644 hints/set-zeroes-in-matrix.md create mode 100644 hints/spiral-matrix.md diff --git a/articles/plus-one.md b/articles/plus-one.md index 242d171fb..4b51b5b0c 100644 --- a/articles/plus-one.md +++ b/articles/plus-one.md @@ -139,7 +139,7 @@ class Solution { --- -## 2. Iteration +## 2. Iteration - I ::tabs-start @@ -148,7 +148,7 @@ class Solution: def plusOne(self, digits: List[int]) -> List[int]: one = 1 i = 0 - digits = digits[::-1] + digits.reverse() while one: if i < len(digits): @@ -161,7 +161,9 @@ class Solution: digits.append(one) one = 0 i += 1 - return digits[::-1] + + digits.reverse() + return digits ``` ```java @@ -344,11 +346,11 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Space complexity: $O(1)$ or $O(n)$ depending on the language. --- -## 3. Iteration (Optimal) +## 3. Iteration - II ::tabs-start @@ -479,4 +481,4 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ \ No newline at end of file +* Space complexity: $O(n)$ \ No newline at end of file diff --git a/articles/set-zeroes-in-matrix.md b/articles/set-zeroes-in-matrix.md index 905883b5c..5a265e1f5 100644 --- a/articles/set-zeroes-in-matrix.md +++ b/articles/set-zeroes-in-matrix.md @@ -212,7 +212,7 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(m * n)$ +* Time complexity: $O((m * n) * (m + n))$ * Space complexity: $O(m * n)$ > Where $m$ is the number of rows and $n$ is the number of columns. diff --git a/articles/spiral-matrix.md b/articles/spiral-matrix.md index b9719b928..7935874fc 100644 --- a/articles/spiral-matrix.md +++ b/articles/spiral-matrix.md @@ -203,7 +203,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(m * n)$ -* Space complexity: $O(min(m, n))$ +* Space complexity: + * $O(min(m, n))$ space for recursion stack. + * $O(m * n)$ space for the output list. > Where $m$ is the number of rows and $n$ is the number of columns. @@ -467,7 +469,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(m * n)$ -* Space complexity: $O(1)$ +* Space complexity: + * $O(1)$ extra space. + * $O(m * n)$ space for the output list. > Where $m$ is the number of rows and $n$ is the number of columns. @@ -655,6 +659,8 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(m * n)$ -* Space complexity: $O(1)$ +* Space complexity: + * $O(1)$ extra space. + * $O(m * n)$ space for the output list. > Where $m$ is the number of rows and $n$ is the number of columns. \ No newline at end of file diff --git a/hints/non-cyclical-number.md b/hints/non-cyclical-number.md new file mode 100644 index 000000000..6672425b9 --- /dev/null +++ b/hints/non-cyclical-number.md @@ -0,0 +1,23 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(logn) time and O(logn) space, where n is the given integer. +

+
+ +
+
+ Hint 1 +

+ Create a helper function that returns the sum of the squares of a number's digits. Then, simulate the given process. If we reach 1, return true. However, we may get stuck in a cycle if a number is processed more than once. What data structure can be used to detect if a number has already been processed? +

+
+ +
+
+ Hint 2 +

+ We can use a hash set to detect if a number has already been processed. At each step, we update n with the return value of the helper function. If the result is 1, we return true. If n is already in the set, we return false. Otherwise, we add n to the hash set and continue. +

+
\ No newline at end of file diff --git a/hints/plus-one.md b/hints/plus-one.md new file mode 100644 index 000000000..ec62b78c9 --- /dev/null +++ b/hints/plus-one.md @@ -0,0 +1,23 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(n) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ There are two cases when adding 1 to a number. If the rightmost digit is less than 9, we simply increment it. Otherwise, we set it to 0 and apply the same process to the preceding digit. +

+
+ +
+
+ Hint 2 +

+ We iterate through the given digits from right to left using index i. If the current digit is less than 9, we increment it and return the array. Otherwise, we set the digit to 0 and continue. If the loop completes without returning, we insert 1 at the beginning of the array and return it. +

+
\ No newline at end of file diff --git a/hints/rotate-matrix.md b/hints/rotate-matrix.md new file mode 100644 index 000000000..3f5528d0f --- /dev/null +++ b/hints/rotate-matrix.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n^2) time and O(1) space, where n is the length of the side of the given square matrix. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would use O(n^2) extra space to solve the problem. Can you think of a way to avoid using extra space? Maybe you should consider observing the positions of the elements before rotating and after rotating of the matrix. +

+
+ +
+
+ Hint 2 +

+ We can rotate the matrix in two steps. First, we reverse the matrix vertically, meaning the first row becomes the last, the second row becomes the second last, and so on. Next, we transpose the reversed matrix, meaning rows become columns and columns become rows. How would you transpose the matrix? +

+
+ +
+
+ Hint 3 +

+ Since the given matrix is a square matrix, we only need to iterate over the upper triangular part, meaning the right upper portion of the main diagonal. In this way, we can transpose a matrix. +

+
\ No newline at end of file diff --git a/hints/set-zeroes-in-matrix.md b/hints/set-zeroes-in-matrix.md new file mode 100644 index 000000000..663ff919e --- /dev/null +++ b/hints/set-zeroes-in-matrix.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(m*n) time and O(1) space, where m is the number of rows and n is the number of columns in the given matrix. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would iterate through the given matrix and update the corresponding row and column in a new matrix on the fly. This would result in an O((m*n)*(m+n)) time and O(m*n) space solution. Can you think of a better way? Maybe you should consider using a single variable for a row and a single variable for a column instead of updating entire row and column. +

+
+ +
+
+ Hint 2 +

+ A better approach is to use O(m+n) boolean arrays. We iterate through the matrix, and when encountering a zero, we mark the respective row and column as true. In the second iteration, we set a cell to 0 if its corresponding row or column is marked true. Can you think of a way to optimize the space further? +

+
+ +
+
+ Hint 3 +

+ We can use the topmost row and leftmost column of the matrix as boolean arrays by marking 0 instead of true. However, since they overlap at one cell, we use a single variable to track the top row separately. We then iterate through the matrix and mark zeros accordingly. +

+
+ +
+
+ Hint 4 +

+ In the second iteration, we update all cells that are not part of the top row or left column accordingly. After making the necessary changes, we check the top-leftmost cell and update the corresponding column. Finally, we check the extra variable and update the top row accordingly. +

+
\ No newline at end of file diff --git a/hints/spiral-matrix.md b/hints/spiral-matrix.md new file mode 100644 index 000000000..2b8eced0a --- /dev/null +++ b/hints/spiral-matrix.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(m*n) time and O(1) extra space, where m is the number of rows and n is the number of columns in the given matrix. +

+
+ +
+
+ Hint 1 +

+ Try to simulate the process as described in the problem. Think in terms of matrix layers, starting from the outermost boundaries and moving inward. Can you determine an efficient way to implement this? +

+
+ +
+
+ Hint 2 +

+ Each boundary consists of four parts: the top row, right column, bottom row, and left column, which follow the spiral order and act as four pointers. For each layer, the top pointer increments by one, the right pointer decrements by one, the left pointer increments by one, and the bottom pointer decrements by one. +

+
+ +
+
+ Hint 3 +

+ At each layer, four loops traverse the matrix: one moves left to right along the top row, another moves top to bottom along the right column, the next moves right to left along the bottom row, and the last moves bottom to top along the left column. This process generates the spiral order. +

+
\ No newline at end of file From f67f984f1f0a900c1d17fc9baf2f266bcef68951 Mon Sep 17 00:00:00 2001 From: Sri Hari Date: Sun, 9 Mar 2025 23:07:42 +0530 Subject: [PATCH 4/5] Batch-6/Neetcode-150/Added-hints --- articles/counting-bits.md | 26 +++++++++++++------ articles/string-encode-and-decode.md | 8 +++--- hints/count-squares.md | 39 ++++++++++++++++++++++++++++ hints/counting-bits.md | 31 ++++++++++++++++++++++ hints/missing-number.md | 39 ++++++++++++++++++++++++++++ hints/multiply-strings.md | 39 ++++++++++++++++++++++++++++ hints/number-of-one-bits.md | 31 ++++++++++++++++++++++ hints/pow-x-n.md | 39 ++++++++++++++++++++++++++++ hints/reverse-bits.md | 31 ++++++++++++++++++++++ hints/reverse-integer.md | 31 ++++++++++++++++++++++ hints/single-number.md | 39 ++++++++++++++++++++++++++++ hints/string-encode-and-decode.md | 2 +- hints/sum-of-two-integers.md | 39 ++++++++++++++++++++++++++++ 13 files changed, 381 insertions(+), 13 deletions(-) create mode 100644 hints/count-squares.md create mode 100644 hints/counting-bits.md create mode 100644 hints/missing-number.md create mode 100644 hints/multiply-strings.md create mode 100644 hints/number-of-one-bits.md create mode 100644 hints/pow-x-n.md create mode 100644 hints/reverse-bits.md create mode 100644 hints/reverse-integer.md create mode 100644 hints/single-number.md create mode 100644 hints/sum-of-two-integers.md diff --git a/articles/counting-bits.md b/articles/counting-bits.md index 514bef953..ac8139aba 100644 --- a/articles/counting-bits.md +++ b/articles/counting-bits.md @@ -124,8 +124,10 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Time complexity: $O(n \log n)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output array. --- @@ -248,8 +250,10 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Time complexity: $O(n \log n)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output array. --- @@ -338,8 +342,10 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Time complexity: $O(n \log n)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output array. --- @@ -470,7 +476,9 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output array. --- @@ -567,4 +575,6 @@ class Solution { ### Time & Space Complexity * Time complexity: $O(n)$ -* Space complexity: $O(1)$ \ No newline at end of file +* Space complexity: + * $O(1)$ extra space. + * $O(n)$ space for the output array. \ No newline at end of file diff --git a/articles/string-encode-and-decode.md b/articles/string-encode-and-decode.md index c0cad7e12..b60d5115e 100644 --- a/articles/string-encode-and-decode.md +++ b/articles/string-encode-and-decode.md @@ -286,8 +286,8 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(m)$ for $encode()$ and $decode()$. -* Space complexity: $O(n)$ for $encode()$ and $decode()$. +* Time complexity: $O(m)$ for each $encode()$ and $decode()$ function calls. +* Space complexity: $O(m + n)$ for each $encode()$ and $decode()$ function calls. > Where $m$ is the sum of lengths of all the strings and $n$ is the number of strings. @@ -510,7 +510,7 @@ class Solution { ### Time & Space Complexity -* Time complexity: $O(m)$ for $encode()$ and $decode()$. -* Space complexity: $O(1)$ for $encode()$ and $decode()$. +* Time complexity: $O(m)$ for each $encode()$ and $decode()$ function calls. +* Space complexity: $O(m + n)$ for each $encode()$ and $decode()$ function calls. > Where $m$ is the sum of lengths of all the strings and $n$ is the number of strings. \ No newline at end of file diff --git a/hints/count-squares.md b/hints/count-squares.md new file mode 100644 index 000000000..69b289bdb --- /dev/null +++ b/hints/count-squares.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(1) time for each add() call, O(n) time for each count() call, and O(n) space, where n is the total number of points. +

+
+ +
+
+ Hint 1 +

+ Initially, we can store the points in a global list for the add() call. For the count() call, a brute force approach would use three nested loops to check other points except for the query point, resulting in an O(n^3) time solution. Can you think of a better way? Maybe you should consider the observation that can be drawn from the diagonal of a square. +

+
+ +
+
+ Hint 2 +

+ In a square's diagonal, the absolute difference between the x-coordinates is equal to the absolute difference between the y-coordinates of the two endpoints, and neither difference can be zero. Using these two points, we can determine the other diagonal endpoints. +

+
+ +
+
+ Hint 3 +

+ We store points in a hash map instead of a list for O(1) lookups, treating duplicate points as one while tracking their frequencies. For the count() function, we iterate through points that, along with the query point, can form a diagonal. Let the query point be (qx, qy) and the other point be (x, y), ensuring they form a diagonal. What could be the other two points? Maybe you should consider the points forming a right-to-left diagonal, treating (qx, qy) as the top-right corner. +

+
+ +
+
+ Hint 4 +

+ The other two points are point1 (x, qy) and point2 (qx, y). For counting, we simply add count of point1 * count of point2 to the result res. +

+
\ No newline at end of file diff --git a/hints/counting-bits.md b/hints/counting-bits.md new file mode 100644 index 000000000..45188a165 --- /dev/null +++ b/hints/counting-bits.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(n) space, where n is the given integer. +

+
+ +
+
+ Hint 1 +

+ A straightforward approach would be to iterate from 0 to n using i, and for each i, iterate through its bits to count the number of set bits. This would result in an O(n \log n) approach. Can you think of a better way? Maybe you should try identifying a pattern by observing the bitwise representation of consecutive numbers. +

+
+ +
+
+ Hint 2 +

+ For example, to compute set bits for 7, add 1 to the count of set bits in 3, which was previously computed by adding 1 to the count of set bits in 1. Observing the pattern, for numbers less than 4, add 1 to the count from two positions before. For numbers less than 8, add 1 to the count from four positions before. Can you derive a dynamic programming relation from this? +

+
+ +
+
+ Hint 3 +

+ We find an offset for the current number based on the number before offset positions. The dynamic programming relation is dp[i] = 1 + dp[i - offset], where dp[i] represents the number of set bits in i. The offset starts at 1 and updates when encountering a power of 2. To simplify the power of 2 check, if offset * 2 equals the current number, we update offset to the current number. +

+
\ No newline at end of file diff --git a/hints/missing-number.md b/hints/missing-number.md new file mode 100644 index 000000000..2d5f6908b --- /dev/null +++ b/hints/missing-number.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would iterate through the range of numbers from 0 to n, checking if each number is present in the given array. If a number is missing, it is returned. This results in an O(n^2) solution. Can you think of a better way? Maybe a data structure could help optimize this process. +

+
+ +
+
+ Hint 2 +

+ We can use a hash set by inserting the given array elements into it. Then, we iterate through the range of numbers from 0 to n and use the hash set for O(1) lookups to find the missing number. Can you think of a way to further optimize this? Maybe a bitwise operator could be useful. +

+
+ +
+
+ Hint 3 +

+ We can use bitwise XOR. When two identical numbers are XORed, the result is 0. Using this property, we can efficiently find the missing number. +

+
+ +
+
+ Hint 4 +

+ We first compute the bitwise XOR of numbers from 0 to n. Then, we iterate through the array and XOR its elements as well. The missing number remains in the final XOR result since all other numbers appear twice—once in the range and once in the array—while the missing number is XORed only once. +

+
\ No newline at end of file diff --git a/hints/multiply-strings.md b/hints/multiply-strings.md new file mode 100644 index 000000000..193cc4c27 --- /dev/null +++ b/hints/multiply-strings.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(m*n) time and O(m+n) space, where m is the length of the string num1 and n is the length of the string num2. +

+
+ +
+
+ Hint 1 +

+ Implement a helper function that takes two number strings and returns their sum. Ensure that the length of num1 is greater than num2, swapping them if necessary. Can you think of a way to multiply the strings? Maybe you should first consider basic multiplication, where num1 is multiplied by each digit of num2. +

+
+ +
+
+ Hint 2 +

+ When multiplying num1 with each digit of num2, we iterate through num2 in reverse order. Based on the digit's position, we pad zeros to the multiplication result accordingly—no padding for the last digit, one zero for the second last, and so on. What should be the next step after each multiplication? Maybe you should implement a helper function to handle this. +

+
+ +
+
+ Hint 3 +

+ We implement a helper function that takes num1, a digit, and a variable zeroes, returning the multiplication result with zeroes padded at the end. A global string res stores the final result. +

+
+ +
+
+ Hint 4 +

+ In the main function, we iterate through num2 in reverse order, calling the helper function to multiply num1 with the current digit and append the appropriate number of padding zeros. We then call another helper function that takes this multiplication result and the global result string res, adds them, and updates res. +

+
\ No newline at end of file diff --git a/hints/number-of-one-bits.md b/hints/number-of-one-bits.md new file mode 100644 index 000000000..788499e2b --- /dev/null +++ b/hints/number-of-one-bits.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(1) time and O(1) space. +

+
+ +
+
+ Hint 1 +

+ The given integer is a 32-bit integer. Can you think of using bitwise operators to iterate through its bits? Maybe you should consider iterating 32 times. +

+
+ +
+
+ Hint 2 +

+ We iterate 32 times (0 to 31) using index i. The expression (1 << i) creates a bitmask with a set bit at the i-th position. How can you check whether the i-th bit is set in the given number? Maybe you should consider using the bitwise-AND ("&"). +

+
+ +
+
+ Hint 3 +

+ Since the mask has a set bit at the i-th position and all 0s elsewhere, we can perform a bitwise-AND with n. If n has a set bit at the i-th position, the result is positive; otherwise, it is 0. We increment the global count if the result is positive and return it after the iteration. +

+
\ No newline at end of file diff --git a/hints/pow-x-n.md b/hints/pow-x-n.md new file mode 100644 index 000000000..de42350f0 --- /dev/null +++ b/hints/pow-x-n.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution as good or better than O(logn) time and O(logn) space, where n is the given integer. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would be to iterate linearly up to n, multiplying by x each time to find (x^n). If n is negative, return (1 / (x^n)); otherwise, return (x^n). Can you think of a better way? Maybe a recursive approach would be more efficient. +

+
+ +
+
+ Hint 2 +

+ For example, to calculate 2^6, instead of multiplying 2 six times, we compute 2^3 and square the result. The same logic applies recursively to find 2^3 and further break down the multiplication. What should be the base case for this recursion? Maybe you should consider the term that cannot be further broken down. +

+
+ +
+
+ Hint 3 +

+ In (x^n), if x is 0, we return 0. If n is 0, we return 1, as any number raised to the power of 0 is 1. Otherwise, we compute (x^(n/2)) recursively and square the result. If n is odd, we multiply the final result by x. What should be the logic if n is negative? +

+
+ +
+
+ Hint 4 +

+ We start the recursion with the absolute value of n. After computing the result as res, we return res if n is non-negative; otherwise, we return (1 / res). +

+
\ No newline at end of file diff --git a/hints/reverse-bits.md b/hints/reverse-bits.md new file mode 100644 index 000000000..0eb364ce9 --- /dev/null +++ b/hints/reverse-bits.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(1) time and O(1) space. +

+
+ +
+
+ Hint 1 +

+ Given a 32-bit integer, what is the position of bit i after reversing the bits? Maybe you should observe the bit positions before and after reversal to find a pattern. +

+
+ +
+
+ Hint 2 +

+ After reversing the bits, the bit at position i moves to position 31 - i. Can you use this observation to construct the reversed number efficiently? +

+
+ +
+
+ Hint 3 +

+ We initialize res to 0 and iterate through the bits of the given integer n. We extract the bit at the i-th position using ((n >> i) & 1). If it is 1, we set the corresponding bit in res at position (31 - i) using (res |= (1 << (31 - i))). +

+
\ No newline at end of file diff --git a/hints/reverse-integer.md b/hints/reverse-integer.md new file mode 100644 index 000000000..23b4d9c75 --- /dev/null +++ b/hints/reverse-integer.md @@ -0,0 +1,31 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(1) time and O(1) space. +

+
+ +
+
+ Hint 1 +

+ A straightforward approach would be to convert the given integer to a string, reverse it, convert it back to an integer using a long type, and return 0 if the result exceeds the integer range. Can you think of a better way? +

+
+ +
+
+ Hint 2 +

+ We initially declare the result res as an int with a value of 0. We iterate through the given integer, extracting digits one by one. Before appending a digit to res, we consider multiple cases. Can you determine them? Maybe you should think about overflow. +

+
+ +
+
+ Hint 3 +

+ Let MAX be the maximum positive integer and MIN be the minimum negative integer. We iterate through each digit and check for overflow before updating res. If res > MAX / 10 or res < MIN / 10, return 0. If res == MAX / 10 and the current digit is greater than MAX % 10, return 0. If res == MIN / 10 and the current digit is less than MIN % 10, return 0. Otherwise, append the digit to res and continue. +

+
\ No newline at end of file diff --git a/hints/single-number.md b/hints/single-number.md new file mode 100644 index 000000000..6aab0485f --- /dev/null +++ b/hints/single-number.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(n) time and O(1) space, where n is the size of the input array. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would iterate through the array, checking each element using a nested loop. If a duplicate is found, we continue to the next element; otherwise, the current element is the required number. This results in an O(n^2) solution. Can you think of a better way? Maybe a data structure can help detect duplicates efficiently. +

+
+ +
+
+ Hint 2 +

+ We use a hash set, iterating through the array and adding elements that are not in the set while removing those that are already present. After the iteration, the required number remains in the hash set. This results in an O(n) space solution. Can you further optimize it? Maybe a bitwise operator could be useful here. +

+
+ +
+
+ Hint 3 +

+ Think about the bitwise XOR ("^"). What is the result when two identical numbers are XORed? +

+
+ +
+
+ Hint 4 +

+ When two identical numbers are XORed, they cancel out, resulting in zero. Since every number appears twice except for one, the XOR of the entire array gives the number that appears only once. +

+
\ No newline at end of file diff --git a/hints/string-encode-and-decode.md b/hints/string-encode-and-decode.md index 39083d5be..82bc8fbcb 100644 --- a/hints/string-encode-and-decode.md +++ b/hints/string-encode-and-decode.md @@ -2,7 +2,7 @@
Recommended Time & Space Complexity

- You should aim for a solution with O(m) time and O(1) space for each encode() and decode() call, where m is the sum of lengths of all the strings. + You should aim for a solution with O(m) time for each encode() and decode() call and O(m+n) space, where m is the sum of lengths of all the strings and n is the number of strings.

diff --git a/hints/sum-of-two-integers.md b/hints/sum-of-two-integers.md new file mode 100644 index 000000000..340f197d0 --- /dev/null +++ b/hints/sum-of-two-integers.md @@ -0,0 +1,39 @@ +
+
+ Recommended Time & Space Complexity +

+ You should aim for a solution with O(1) time and O(1) space. +

+
+ +
+
+ Hint 1 +

+ A brute force approach would use the addition operator. Can you think of a way to perform addition without using it? Maybe you should consider solving this using bit manipulation. +

+
+ +
+
+ Hint 2 +

+ We can use the bitwise XOR operator to perform addition. If both a and b have 1 at the same bit position, the sum at that position is 0, and a carry of 1 is generated. If the bits are different, the sum at that position is 1. Additionally, we account for the carry from the previous step in the next iteration. +

+
+ +
+
+ Hint 3 +

+ We iterate bit by bit from 0 to 31 since the given integers are 32-bit. We track the carry, initially set to 0, and initialize the result as res. During iteration, the XOR of the bits at the i-th position of both integers and the carry determines the current bit of res. How can you handle negative numbers? +

+
+ +
+
+ Hint 4 +

+ To handle negative numbers, if the final result exceeds the maximum positive 32-bit integer, it means the number should be negative. We adjust it using bitwise operations: flipping the bits with res ^ ((2 ^ 32) - 1) and applying ~ to restore the correct two’s complement representation. This ensures the result correctly represents signed 32-bit integers. +

+
\ No newline at end of file From 1f63b92e7b542bb040f280dce2fa5b3787b68139 Mon Sep 17 00:00:00 2001 From: Sri Hari Date: Mon, 10 Mar 2025 07:12:34 +0530 Subject: [PATCH 5/5] Batch-6/Neetcode-150/Added-hints --- articles/subarrays-with-k-different-integers.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/articles/subarrays-with-k-different-integers.md b/articles/subarrays-with-k-different-integers.md index 26328205c..a2e24a470 100644 --- a/articles/subarrays-with-k-different-integers.md +++ b/articles/subarrays-with-k-different-integers.md @@ -246,7 +246,7 @@ class Solution { --- -## 3. Slidingt Window (One Pass) - I +## 3. Sliding Window (One Pass) - I ::tabs-start