Minimum Operations to Make Array Equal II

Problem

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.

Example 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.

Example 2:

Input: nums1 = [3,8,5,2], nums2 = [2,4,1,6], k = 1
Output: -1
Explanation: It can be proved that it is impossible to make the two arrays equal.

Constraints:

n == nums1.length == nums2.length
2 <= n <= 105
0 <= nums1[i], nums2[j] <= 109
0 <= k <= 105

Solution (Java)

class Solution {
    public long minOperations(int[] nums1, int[] nums2, int k) {

    int n = nums1.length;
    long[] diff = new long[n];
    for (int i = 0; i < n; i++) {
        diff[i] = nums2[i] - nums1[i];
    }
    if(k==0){
        for(int i=0;i<n;i++)
            if(diff[i]!=0)return -1;
        return 0;
    }
    long ans = 0;
    long sum=0;
    for (int i = 0; i < n; i++) {
        sum+=diff[i];
        if (diff[i] % k != 0) {
            return -1;
        }
        ans += Math.abs(diff[i] / k);
    }

    return sum==0? ans/2:-1;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).