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Commite433db1

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Add solution #2435
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‎README.md

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#1,945 LeetCode solutions in JavaScript
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#1,946 LeetCode solutions in JavaScript
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[https://leetcodejavascript.com](https://leetcodejavascript.com)
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2432|[The Employee That Worked on the Longest Task](./solutions/2432-the-employee-that-worked-on-the-longest-task.js)|Easy|
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2433|[Find The Original Array of Prefix Xor](./solutions/2433-find-the-original-array-of-prefix-xor.js)|Medium|
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2434|[Using a Robot to Print the Lexicographically Smallest String](./solutions/2434-using-a-robot-to-print-the-lexicographically-smallest-string.js)|Medium|
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2435|[Paths in Matrix Whose Sum Is Divisible by K](./solutions/2435-paths-in-matrix-whose-sum-is-divisible-by-k.js)|Hard|
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2444|[Count Subarrays With Fixed Bounds](./solutions/2444-count-subarrays-with-fixed-bounds.js)|Hard|
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2460|[Apply Operations to an Array](./solutions/2460-apply-operations-to-an-array.js)|Easy|
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2462|[Total Cost to Hire K Workers](./solutions/2462-total-cost-to-hire-k-workers.js)|Medium|
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/**
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* 2435. Paths in Matrix Whose Sum Is Divisible by K
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* https://leetcode.com/problems/paths-in-matrix-whose-sum-is-divisible-by-k/
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* Difficulty: Hard
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*
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* You are given a 0-indexed m x n integer matrix grid and an integer k. You are currently at
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* position (0, 0) and you want to reach position (m - 1, n - 1) moving only down or right.
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*
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* Return the number of paths where the sum of the elements on the path is divisible by k.
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* Since the answer may be very large, return it modulo 109 + 7.
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*/
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/**
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*@param {number[][]} grid
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*@param {number} k
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*@return {number}
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*/
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varnumberOfPaths=function(grid,k){
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constrows=grid.length;
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constcols=grid[0].length;
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constmodulo=1e9+7;
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constdp=Array.from({length:rows},()=>{
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returnArray.from({length:cols},()=>Array(k).fill(0));
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});
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dp[0][0][grid[0][0]%k]=1;
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for(letrow=0;row<rows;row++){
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for(letcol=0;col<cols;col++){
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constcurrentValue=grid[row][col]%k;
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for(letsum=0;sum<k;sum++){
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if(row>0){
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constprevSum=(sum-currentValue+k)%k;
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dp[row][col][sum]=(dp[row][col][sum]+dp[row-1][col][prevSum])%modulo;
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}
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if(col>0){
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constprevSum=(sum-currentValue+k)%k;
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dp[row][col][sum]=(dp[row][col][sum]+dp[row][col-1][prevSum])%modulo;
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}
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}
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}
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}
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returndp[rows-1][cols-1][0];
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};

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