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701 changes: 701 additions & 0 deletions articles/constrained-subsequence-sum.md

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543 changes: 543 additions & 0 deletions articles/find-the-kth-largest-integer-in-the-array.md

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387 changes: 387 additions & 0 deletions articles/find-the-longest-valid-obstacle-course-at-each-position.md
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## 1. Dynamic Programming (Top-Down)

::tabs-start

```python
class Solution:
def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
n = len(obstacles)
dp = [[-1] * (n + 1) for _ in range(n)]

def dfs(i, prev):
if i < 0:
return 0
if dp[i][prev] != -1:
return dp[i][prev]

res = dfs(i - 1, prev)
if prev == n or obstacles[prev] >= obstacles[i]:
res = max(res, 1 + dfs(i - 1, i))
dp[i][prev] = res
return res

dfs(n - 1, n)
return [1] + [1 + dp[i - 1][i] for i in range(1, n)]
```

```java
public class Solution {
private int[][] dp;

public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
int n = obstacles.length;
this.dp = new int[n][n + 1];
for (int[] row : dp) {
Arrays.fill(row, -1);
}

dfs(n - 1, n, obstacles);

int[] res = new int[n];
res[0] = 1;
for (int i = 1; i < n; i++) {
res[i] = 1 + dp[i - 1][i];
}
return res;
}

private int dfs(int i, int prev, int[] obstacles) {
if (i < 0) {
return 0;
}
if (dp[i][prev] != -1) {
return dp[i][prev];
}

int res = dfs(i - 1, prev, obstacles);
if (prev == obstacles.length || obstacles[prev] >= obstacles[i]) {
res = Math.max(res, 1 + dfs(i - 1, i, obstacles));
}
return dp[i][prev] = res;
}
}
```

```cpp
class Solution {
public:
vector<vector<int>> dp;

vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
int n = obstacles.size();
this->dp = vector<vector<int>>(n, vector<int>(n + 1, -1));

dfs(n - 1, n, obstacles);

vector<int> res(n, 1);
for (int i = 1; i < n; i++) {
res[i] = 1 + dp[i - 1][i];
}
return res;
}

private:
int dfs(int i, int prev, vector<int>& obstacles) {
if (i < 0) {
return 0;
}
if (dp[i][prev] != -1) {
return dp[i][prev];
}

int res = dfs(i - 1, prev, obstacles);
if (prev == obstacles.size() || obstacles[prev] >= obstacles[i]) {
res = max(res, 1 + dfs(i - 1, i, obstacles));
}
return dp[i][prev] = res;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} obstacles
* @return {number[]}
*/
longestObstacleCourseAtEachPosition(obstacles) {
const n = obstacles.length;
const dp = Array.from({ length: n }, () => new Array(n + 1).fill(-1));

const dfs = (i, prev) => {
if (i < 0) {
return 0;
}
if (dp[i][prev] !== -1) {
return dp[i][prev];
}

let res = dfs(i - 1, prev);
if (prev === n || obstacles[prev] >= obstacles[i]) {
res = Math.max(res, 1 + dfs(i - 1, i));
}
dp[i][prev] = res;
return res;
};

dfs(n - 1, n);

const res = new Array(n).fill(1);
for (let i = 1; i < n; i++) {
res[i] = 1 + dp[i - 1][i];
}
return res;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n ^ 2)$
* Space complexity: $O(n ^ 2)$

---

## 2. Dynamic Programming (Binary Search) - I

::tabs-start

```python
class Solution:
def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
res = []
dp = [10**8] * (len(obstacles) + 1)

for num in obstacles:
index = bisect.bisect(dp, num)
res.append(index + 1)
dp[index] = num

return res
```

```java
public class Solution {
public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
int n = obstacles.length;
int[] res = new int[n];
int[] dp = new int[n + 1];
Arrays.fill(dp, (int) 1e8);

for (int i = 0; i < n; i++) {
int index = upperBound(dp, obstacles[i]);
res[i] = index + 1;
dp[index] = obstacles[i];
}

return res;
}

private int upperBound(int[] dp, int target) {
int left = 0, right = dp.length;
while (left < right) {
int mid = left + (right - left) / 2;
if (dp[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}
```

```cpp
class Solution {
public:
vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
int n = obstacles.size();
vector<int> res(n);
vector<int> dp(n + 1, 1e8);

for (int i = 0; i < n; i++) {
int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
res[i] = index + 1;
dp[index] = obstacles[i];
}

return res;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} obstacles
* @return {number[]}
*/
longestObstacleCourseAtEachPosition(obstacles) {
let n = obstacles.length;
let res = new Array(n).fill(0);
let dp = new Array(n + 1).fill(1e8);

const upperBound = (dp, target) => {
let left = 0, right = dp.length;
while (left < right) {
let mid = Math.floor((left + right) / 2);
if (dp[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};

for (let i = 0; i < n; i++) {
let index = upperBound(dp, obstacles[i]);
res[i] = index + 1;
dp[index] = obstacles[i];
}

return res;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n \log n)$
* Space complexity: $O(n)$

---

## 3. Dynamic Programming (Binary Search) - II

::tabs-start

```python
class Solution:
def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
res = []
dp = []

for num in obstacles:
index = bisect.bisect_right(dp, num)
res.append(index + 1)

if index == len(dp):
dp.append(num)
else:
dp[index] = num

return res
```

```java
public class Solution {
public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
int n = obstacles.length;
int[] res = new int[n];
List<Integer> dp = new ArrayList<>();

for (int i = 0; i < n; i++) {
int index = upperBound(dp, obstacles[i]);
res[i] = index + 1;

if (index == dp.size()) {
dp.add(obstacles[i]);
} else {
dp.set(index, obstacles[i]);
}
}

return res;
}

private int upperBound(List<Integer> dp, int target) {
int left = 0, right = dp.size();
while (left < right) {
int mid = left + (right - left) / 2;
if (dp.get(mid) > target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}
```

```cpp
class Solution {
public:
vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
int n = obstacles.size();
vector<int> res(n);
vector<int> dp;

for (int i = 0; i < n; i++) {
int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
res[i] = index + 1;

if (index == dp.size()) {
dp.push_back(obstacles[i]);
} else {
dp[index] = obstacles[i];
}
}

return res;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} obstacles
* @return {number[]}
*/
longestObstacleCourseAtEachPosition(obstacles) {
let n = obstacles.length;
let res = new Array(n).fill(0);
let dp = [];

const upperBound = (dp, target) => {
let left = 0, right = dp.length;
while (left < right) {
let mid = Math.floor((left + right) / 2);
if (dp[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};

for (let i = 0; i < n; i++) {
let index = upperBound(dp, obstacles[i]);
res[i] = index + 1;

if (index === dp.length) {
dp.push(obstacles[i]);
} else {
dp[index] = obstacles[i];
}
}

return res;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n \log n)$
* Space complexity: $O(n)$
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