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2541. Minimum Operations to Make Array Equal II (Medium)

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

Companies: Walmart Labs

Related Topics:
Array, Math, Greedy

Similar Questions:

• Minimum Operations to Make Array Equal (Medium)
• Minimum Number of Operations to Make Arrays Similar (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/minimum-operations-to-make-array-equal-ii
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    long long minOperations(vector<int>& A, vector<int>& B, int k) {
        if (k == 0) return A == B ? 0 : -1;
        long long ans = 0, diffSum = 0;
        for (int i = 0; i < A.size(); ++i) {
            if (A[i] == B[i]) continue;
            long long d = (long long)A[i] - B[i];
            if (abs(d) % k) return -1;
            diffSum += d;
            if (d > 0) ans += abs(d) / k;
        }
        return diffSum == 0 ? ans : -1;
    }
};