🤖 leetcode 2541. Minimum Operations to Make Array Equal II | medium | greedy | javascript

🗒️ Problems

2541. Minimum Operations to Make Array Equal II - Leetcode

You are given two integer arrays nums1 and nums2 of equal length n and an integer k.
You can perform the following operation on nums1:

Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k.
In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k.
nums1 is said to be equal to nums2 if for all indices i
such that 0 <= i < n, nums1[i] == nums2[i].

Return the minimum number of operations required to make nums1 equal to nums2.
If it is impossible to make them equal, return -1.

Input: nums1 = [4,3,1,4], nums2 = [1,3,7,1], k = 3
Output: 2
Explanation: In 2 operations, we can transform nums1 to nums2.
1st operation: i = 2, j = 0. After applying the operation, nums1 = [1,3,4,4].
2nd operation: i = 2, j = 3. After applying the operation, nums1 = [1,3,7,1].
One can prove that it is impossible to make arrays equal in fewer operations.

Constraints:

n == nums1.length == nums2.length
2 <= n <= 10^5
0 <= nums1[i], nums2[j] <= 10^9
0 <= k <= 10^5

🤔 Understand problem

• We will change nums1[] to be equal to nums2[] only by adding or subtracting k.

🤦‍♂️ First attempt

• How to solve ?
• This kind of Greedy or Math problem, it's difficult to come up with the solution at first.

🥳 Think differently

• If the diff between nums1[i] and nums2[i] is not the multiple of k, it's impossible to make the two arrays equal.
• First check how many addings and subtractions are necessary to make the two arrays equal.
  • If there are same number of addings and subtractions, it's possible case.
  • If not, it's impossible case.

🔥 My Solution

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @param {number} k
 * @return {number}
 */
var minOperations = function (nums1, nums2, k) {
  const N = nums1.length;

  let pos = 0;
  let nes = 0;

  for (let i = 0; i < N; i += 1) {
    if (nums1[i] === nums2[i]) continue;
    if (Math.abs(nums1[i] - nums2[i]) % k !== 0) return -1;

    if (nums1[i] > nums2[i]) {
      pos += (nums1[i] - nums2[i]) / k;
    } else if (nums1[i] < nums2[i]) {
      nes += (nums2[i] - nums1[i]) / k;
    }
  }

  return pos === nes ? pos : -1;
};

🙆‍♂️ Time complexity: [object Object]