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‎src/main/java/g3201_3300/s3216_lexicographically_smallest_string_after_a_swap/Solution.java

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package g3201_3300.s3216_lexicographically_smallest_string_after_a_swap;
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// #Easy #String #Greedy #2024_07_18_Time_1_ms_(100.00%)_Space_43.3_MB_(20.48%)
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public class Solution {
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public String getSmallestString(String s) {
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final char[] arr = s.toCharArray();
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for (int i = 1; i < arr.length; i++) {
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if (arr[i - 1] % 2 == arr[i] % 2 && arr[i - 1] > arr[i]) {
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final char temp = arr[i];
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arr[i] = arr[i - 1];
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arr[i - 1] = temp;
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break;
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}
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}
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return String.valueOf(arr);
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}
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}

‎src/main/java/g3201_3300/s3216_lexicographically_smallest_string_after_a_swap/readme.md

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3216\. Lexicographically Smallest String After a Swap
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Easy
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Given a string `s` containing only digits, return the lexicographically smallest string that can be obtained after swapping **adjacent** digits in `s` with the same **parity** at most **once**.
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Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, have the same parity, while 6 and 9 do not.
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**Example 1:**
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**Input:** s = "45320"
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**Output:** "43520"
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**Explanation:**
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`s[1] == '5'` and `s[2] == '3'` both have the same parity, and swapping them results in the lexicographically smallest string.
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**Example 2:**
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**Input:** s = "001"
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**Output:** "001"
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**Explanation:**
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There is no need to perform a swap because `s` is already the lexicographically smallest.
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**Constraints:**
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* `2 <= s.length <= 100`
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* `s` consists only of digits.

‎src/main/java/g3201_3300/s3217_delete_nodes_from_linked_list_present_in_array/Solution.java

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package g3201_3300.s3217_delete_nodes_from_linked_list_present_in_array;
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import com_github_leetcode.ListNode;
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// #Medium #Array #Hash_Table #Linked_List #2024_07_18_Time_3_ms_(100.00%)_Space_63.9_MB_(93.81%)
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/**
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* Definition for singly-linked list. public class ListNode { int val; ListNode next; ListNode() {}
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* ListNode(int val) { this.val = val; } ListNode(int val, ListNode next) { this.val = val;
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* this.next = next; } }
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*/
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public class Solution {
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public ListNode modifiedList(int[] nums, ListNode head) {
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int maxv = 0;
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for (int v : nums) {
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maxv = Math.max(maxv, v);
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}
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boolean[] rem = new boolean[maxv + 1];
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for (int v : nums) {
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rem[v] = true;
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}
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ListNode h = new ListNode(0);
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ListNode t = h;
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ListNode p = head;
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while (p != null) {
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if (p.val > maxv || !rem[p.val]) {
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t.next = p;
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t = p;
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}
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p = p.next;
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}
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t.next = null;
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return h.next;
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}
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}

‎src/main/java/g3201_3300/s3217_delete_nodes_from_linked_list_present_in_array/readme.md

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3217\. Delete Nodes From Linked List Present in Array
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Medium
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You are given an array of integers `nums` and the `head` of a linked list. Return the `head` of the modified linked list after **removing** all nodes from the linked list that have a value that exists in `nums`.
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**Example 1:**
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**Input:** nums = [1,2,3], head = [1,2,3,4,5]
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**Output:** [4,5]
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**Explanation:**
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**![](https://assets.leetcode.com/uploads/2024/06/11/linkedlistexample0.png)**
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Remove the nodes with values 1, 2, and 3.
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**Example 2:**
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**Input:** nums = [1], head = [1,2,1,2,1,2]
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**Output:** [2,2,2]
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/06/11/linkedlistexample1.png)
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Remove the nodes with value 1.
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**Example 3:**
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**Input:** nums = [5], head = [1,2,3,4]
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**Output:** [1,2,3,4]
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**Explanation:**
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**![](https://assets.leetcode.com/uploads/2024/06/11/linkedlistexample2.png)**
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No node has value 5.
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>5</sup></code>
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* All elements in `nums` are unique.
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* The number of nodes in the given list is in the range <code>[1, 10<sup>5</sup>]</code>.
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* <code>1 <= Node.val <= 10<sup>5</sup></code>
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* The input is generated such that there is at least one node in the linked list that has a value not present in `nums`.

‎src/main/java/g3201_3300/s3218_minimum_cost_for_cutting_cake_i/Solution.java

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package g3201_3300.s3218_minimum_cost_for_cutting_cake_i;
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// #Medium #Array #Dynamic_Programming #Sorting #Greedy
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// #2024_07_18_Time_0_ms_(100.00%)_Space_42.4_MB_(32.85%)
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public class Solution {
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public int minimumCost(int ignoredM, int ignoredN, int[] horizontalCut, int[] verticalCut) {
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int sum = 0;
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for (int hc : horizontalCut) {
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sum += hc;
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}
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for (int vc : verticalCut) {
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sum += vc;
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}
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for (int hc : horizontalCut) {
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for (int vc : verticalCut) {
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sum += Math.min(hc, vc);
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}
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}
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return sum;
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}
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}

‎src/main/java/g3201_3300/s3218_minimum_cost_for_cutting_cake_i/readme.md

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3218\. Minimum Cost for Cutting Cake I
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Medium
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There is an `m x n` cake that needs to be cut into `1 x 1` pieces.
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You are given integers `m`, `n`, and two arrays:
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* `horizontalCut` of size `m - 1`, where `horizontalCut[i]` represents the cost to cut along the horizontal line `i`.
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* `verticalCut` of size `n - 1`, where `verticalCut[j]` represents the cost to cut along the vertical line `j`.
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In one operation, you can choose any piece of cake that is not yet a `1 x 1` square and perform one of the following cuts:
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1. Cut along a horizontal line `i` at a cost of `horizontalCut[i]`.
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2. Cut along a vertical line `j` at a cost of `verticalCut[j]`.
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After the cut, the piece of cake is divided into two distinct pieces.
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The cost of a cut depends only on the initial cost of the line and does not change.
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Return the **minimum** total cost to cut the entire cake into `1 x 1` pieces.
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**Example 1:**
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**Input:** m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5]
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**Output:** 13
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/06/04/ezgifcom-animated-gif-maker-1.gif)
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* Perform a cut on the vertical line 0 with cost 5, current total cost is 5.
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* Perform a cut on the horizontal line 0 on `3 x 1` subgrid with cost 1.
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* Perform a cut on the horizontal line 0 on `3 x 1` subgrid with cost 1.
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* Perform a cut on the horizontal line 1 on `2 x 1` subgrid with cost 3.
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* Perform a cut on the horizontal line 1 on `2 x 1` subgrid with cost 3.
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The total cost is `5 + 1 + 1 + 3 + 3 = 13`.
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**Example 2:**
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**Input:** m = 2, n = 2, horizontalCut = [7], verticalCut = [4]
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**Output:** 15
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**Explanation:**
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* Perform a cut on the horizontal line 0 with cost 7.
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* Perform a cut on the vertical line 0 on `1 x 2` subgrid with cost 4.
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* Perform a cut on the vertical line 0 on `1 x 2` subgrid with cost 4.
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The total cost is `7 + 4 + 4 = 15`.
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**Constraints:**
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* `1 <= m, n <= 20`
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* `horizontalCut.length == m - 1`
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* `verticalCut.length == n - 1`
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* <code>1 <= horizontalCut[i], verticalCut[i] <= 10<sup>3</sup></code>

‎src/main/java/g3201_3300/s3219_minimum_cost_for_cutting_cake_ii/Solution.java

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package g3201_3300.s3219_minimum_cost_for_cutting_cake_ii;
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// #Hard #Array #Sorting #Greedy #2024_07_18_Time_3_ms_(100.00%)_Space_62.6_MB_(25.82%)
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public class Solution {
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private static final int N = 1001;
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private static final int[] HORIZONTAL_COUNTS = new int[N];
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private static final int[] VERTICAL_COUNTS = new int[N];
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public long minimumCost(int m, int n, int[] horizontalCut, int[] verticalCut) {
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int max = 0;
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for (int x : horizontalCut) {
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if (x > max) {
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max = x;
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}
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HORIZONTAL_COUNTS[x]++;
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}
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for (int x : verticalCut) {
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if (x > max) {
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max = x;
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}
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VERTICAL_COUNTS[x]++;
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}
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long ans = 0;
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int horizontalCount = 1;
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int verticalCount = 1;
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for (int x = max; x > 0; x--) {
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ans += (long) HORIZONTAL_COUNTS[x] * x * horizontalCount;
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verticalCount += HORIZONTAL_COUNTS[x];
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HORIZONTAL_COUNTS[x] = 0;
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ans += (long) VERTICAL_COUNTS[x] * x * verticalCount;
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horizontalCount += VERTICAL_COUNTS[x];
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VERTICAL_COUNTS[x] = 0;
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}
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return ans;
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}
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}

‎src/main/java/g3201_3300/s3219_minimum_cost_for_cutting_cake_ii/readme.md

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3219\. Minimum Cost for Cutting Cake II
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Hard
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There is an `m x n` cake that needs to be cut into `1 x 1` pieces.
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You are given integers `m`, `n`, and two arrays:
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* `horizontalCut` of size `m - 1`, where `horizontalCut[i]` represents the cost to cut along the horizontal line `i`.
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* `verticalCut` of size `n - 1`, where `verticalCut[j]` represents the cost to cut along the vertical line `j`.
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In one operation, you can choose any piece of cake that is not yet a `1 x 1` square and perform one of the following cuts:
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1. Cut along a horizontal line `i` at a cost of `horizontalCut[i]`.
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2. Cut along a vertical line `j` at a cost of `verticalCut[j]`.
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After the cut, the piece of cake is divided into two distinct pieces.
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The cost of a cut depends only on the initial cost of the line and does not change.
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Return the **minimum** total cost to cut the entire cake into `1 x 1` pieces.
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**Example 1:**
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**Input:** m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5]
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**Output:** 13
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/06/04/ezgifcom-animated-gif-maker-1.gif)
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* Perform a cut on the vertical line 0 with cost 5, current total cost is 5.
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* Perform a cut on the horizontal line 0 on `3 x 1` subgrid with cost 1.
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* Perform a cut on the horizontal line 0 on `3 x 1` subgrid with cost 1.
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* Perform a cut on the horizontal line 1 on `2 x 1` subgrid with cost 3.
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* Perform a cut on the horizontal line 1 on `2 x 1` subgrid with cost 3.
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The total cost is `5 + 1 + 1 + 3 + 3 = 13`.
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**Example 2:**
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**Input:** m = 2, n = 2, horizontalCut = [7], verticalCut = [4]
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**Output:** 15
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**Explanation:**
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* Perform a cut on the horizontal line 0 with cost 7.
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* Perform a cut on the vertical line 0 on `1 x 2` subgrid with cost 4.
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* Perform a cut on the vertical line 0 on `1 x 2` subgrid with cost 4.
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The total cost is `7 + 4 + 4 = 15`.
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**Constraints:**
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* <code>1 <= m, n <= 10<sup>5</sup></code>
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* `horizontalCut.length == m - 1`
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* `verticalCut.length == n - 1`
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* <code>1 <= horizontalCut[i], verticalCut[i] <= 10<sup>3</sup></code>

‎src/main/java/g3201_3300/s3220_odd_and_even_transactions/readme.md

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3220\. Odd and Even Transactions
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Medium
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SQL Schema
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Table: `transactions`
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+------------------+------+
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| Column Name | Type |
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+------------------+------+
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| transaction_id | int |
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| amount | int |
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| transaction_date | date |
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+------------------+------+
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The transactions_id column uniquely identifies each row in this table.
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Each row of this table contains the transaction id, amount and transaction date.
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Write a solution to find the **sum of amounts** for **odd** and **even** transactions for each day. If there are no odd or even transactions for a specific date, display as `0`.
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Return _the result table ordered by_ `transaction_date` _in **ascending** order_.
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The result format is in the following example.
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**Example:**
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**Input:**
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`transactions` table:
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+----------------+--------+------------------+
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| transaction_id | amount | transaction_date |
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+----------------+--------+------------------+
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| 1 | 150 | 2024-07-01 |
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| 2 | 200 | 2024-07-01 |
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| 3 | 75 | 2024-07-01 |
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| 4 | 300 | 2024-07-02 |
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| 5 | 50 | 2024-07-02 |
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| 6 | 120 | 2024-07-03 |
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+----------------+--------+------------------+
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**Output:**
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+------------------+---------+----------+
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| transaction_date | odd_sum | even_sum |
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+------------------+---------+----------+
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| 2024-07-01 | 75 | 350 |
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| 2024-07-02 | 0 | 350 |
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| 2024-07-03 | 0 | 120 |
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+------------------+---------+----------+
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**Explanation:**
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* For transaction dates:
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* 2024-07-01:
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* Sum of amounts for odd transactions: 75
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* Sum of amounts for even transactions: 150 + 200 = 350
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* 2024-07-02:
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* Sum of amounts for odd transactions: 0
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* Sum of amounts for even transactions: 300 + 50 = 350
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* 2024-07-03:
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* Sum of amounts for odd transactions: 0
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* Sum of amounts for even transactions: 120
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**Note:** The output table is ordered by `transaction_date` in ascending order.

‎src/main/java/g3201_3300/s3220_odd_and_even_transactions/script.sql

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# Write your MySQL query statement below
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# #Medium #2024_07_18_Time_272_ms_(100.00%)_Space_0B_(100.00%)
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select transaction_date,
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sum(case when amount%2<>0 then amount else 0 end) as odd_sum,
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sum(case when amount%2=0 then amount else 0 end) as even_sum from transactions
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group by transaction_date order by transaction_date;

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