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Commit90acab5

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Add solution #2448
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‎README.md

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#1,955 LeetCode solutions in JavaScript
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#1,956 LeetCode solutions in JavaScript
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[https://leetcodejavascript.com](https://leetcodejavascript.com)
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2444|[Count Subarrays With Fixed Bounds](./solutions/2444-count-subarrays-with-fixed-bounds.js)|Hard|
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2446|[Determine if Two Events Have Conflict](./solutions/2446-determine-if-two-events-have-conflict.js)|Easy|
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2447|[Number of Subarrays With GCD Equal to K](./solutions/2447-number-of-subarrays-with-gcd-equal-to-k.js)|Medium|
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2448|[Minimum Cost to Make Array Equal](./solutions/2448-minimum-cost-to-make-array-equal.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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2467|[Most Profitable Path in a Tree](./solutions/2467-most-profitable-path-in-a-tree.js)|Medium|
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/**
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* 2448. Minimum Cost to Make Array Equal
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* https://leetcode.com/problems/minimum-cost-to-make-array-equal/
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* Difficulty: Hard
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*
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* You are given two 0-indexed arrays nums and cost consisting each of n positive integers.
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*
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* You can do the following operation any number of times:
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* - Increase or decrease any element of the array nums by 1.
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*
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* The cost of doing one operation on the ith element is cost[i].
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*
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* Return the minimum total cost such that all the elements of the array nums become equal.
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*/
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/**
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*@param {number[]} nums
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*@param {number[]} cost
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*@return {number}
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*/
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varminCost=function(nums,cost){
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constpairs=nums.map((num,i)=>[num,cost[i]]).sort((a,b)=>a[0]-b[0]);
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constn=nums.length;
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constprefixSum=Array(n).fill(0);
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constprefixCost=Array(n).fill(0);
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prefixSum[0]=pairs[0][0]*pairs[0][1];
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prefixCost[0]=pairs[0][1];
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for(leti=1;i<n;i++){
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prefixSum[i]=prefixSum[i-1]+pairs[i][0]*pairs[i][1];
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prefixCost[i]=prefixCost[i-1]+pairs[i][1];
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}
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letresult=Infinity;
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for(leti=0;i<n;i++){
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consttarget=pairs[i][0];
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constleftCost=i>0 ?target*prefixCost[i-1]-prefixSum[i-1] :0;
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constrightCost=i<n-1 ?prefixSum[n-1]-prefixSum[i]
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-target*(prefixCost[n-1]-prefixCost[i]) :0;
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result=Math.min(result,leftCost+rightCost);
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}
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returnresult;
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};

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