LeetCode: Equal Sum Arrays With Minimum Number Of Operations Solution

Approach

store the array of changes

• with array of larger sum, calculate the change to 1 for each
• with array of smaller sum, calculate the change to 6 for each

sort the change desc, decrease the init sum diff between two arrays

Implementation

1/**
2 * @param {number[]} nums1
3 * @param {number[]} nums2
4 * @return {number}
5 */
6var minOperations = function (nums1, nums2) {
7  const sum = arr => arr.reduce((acc, el) => acc + el, 0)
8
9  let largerSumArr, smallerSumArr
10  if (sum(nums1) > sum(nums2)) {
11    largerSumArr = [...nums1]
12    smallerSumArr = [...nums2]
13  } else {
14    largerSumArr = [...nums2]
15    smallerSumArr = [...nums1]
16  }
17
18  let res = 0
19  let diff = sum(largerSumArr) - sum(smallerSumArr)
20
21  if (diff === 0) {
22    return 0
23  }
24
25  let changes = [
26    ...largerSumArr.map(el => el - 1),
27    ...smallerSumArr.map(el => 6 - el),
28  ].sort((a, b) => b - a)
29
30  for (const change of changes) {
31    res++
32    diff -= change
33    if (diff <= 0) return res
34  }
35
36  return -1
37}