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Commit20ff7b7

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Added tasks 3216-3220
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packageg3201_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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publicclassSolution {
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publicStringgetSmallestString(Strings) {
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finalchar[]arr =s.toCharArray();
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for (inti =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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finalchartemp =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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returnString.valueOf(arr);
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}
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}
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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.
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packageg3201_3300.s3217_delete_nodes_from_linked_list_present_in_array;
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importcom_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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publicclassSolution {
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publicListNodemodifiedList(int[]nums,ListNodehead) {
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intmaxv =0;
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for (intv :nums) {
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maxv =Math.max(maxv,v);
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}
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boolean[]rem =newboolean[maxv +1];
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for (intv :nums) {
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rem[v] =true;
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}
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ListNodeh =newListNode(0);
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ListNodet =h;
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ListNodep =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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returnh.next;
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}
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}
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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`.
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packageg3201_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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publicclassSolution {
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publicintminimumCost(intignoredM,intignoredN,int[]horizontalCut,int[]verticalCut) {
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intsum =0;
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for (inthc :horizontalCut) {
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sum +=hc;
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}
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for (intvc :verticalCut) {
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sum +=vc;
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}
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for (inthc :horizontalCut) {
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for (intvc :verticalCut) {
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sum +=Math.min(hc,vc);
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}
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}
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returnsum;
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}
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}
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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>
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packageg3201_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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publicclassSolution {
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privatestaticfinalintN =1001;
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privatestaticfinalint[]HORIZONTAL_COUNTS =newint[N];
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privatestaticfinalint[]VERTICAL_COUNTS =newint[N];
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publiclongminimumCost(intm,intn,int[]horizontalCut,int[]verticalCut) {
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intmax =0;
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for (intx :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 (intx :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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longans =0;
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inthorizontalCount =1;
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intverticalCount =1;
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for (intx =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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returnans;
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}
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}
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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>
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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.
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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 else0 end)as odd_sum,
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sum(case when amount%2=0 then amount else0 end)as even_sumfrom transactions
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group by transaction_dateorder by transaction_date;

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