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2541. Minimum Operations to Make Array Equal II

Description

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

Solutions

Solution 1: Single Pass

We use a variable $x$ to record the difference in the number of additions and subtractions, and a variable $ans$ to record the number of operations.

We traverse the array, and for each position $i$, if $k=0$ and $a_i \neq b_i$, then it is impossible to make the two arrays equal, so we return $-1$. Otherwise, if $k \neq 0$, then $a_i - b_i$ must be a multiple of $k$, otherwise it is impossible to make the two arrays equal, so we return $-1$. Next, we update $x$ and $ans$.

Finally, if $x \neq 0$, then it is impossible to make the two arrays equal, so we return $-1$. Otherwise, we return $\frac{ans}{2}$.

The time complexity is $O(n)$, and the space complexity is $O(1)$, where $n$ is the length of the array.

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class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int], k: int) -> int:
        ans = x = 0
        for a, b in zip(nums1, nums2):
            if k == 0:
                if a != b:
                    return -1
                continue
            if (a - b) % k:
                return -1
            y = (a - b) // k
            ans += abs(y)
            x += y
        return -1 if x else ans // 2

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class Solution {
    public long minOperations(int[] nums1, int[] nums2, int k) {
        long ans = 0, x = 0;
        for (int i = 0; i < nums1.length; ++i) {
            int a = nums1[i], b = nums2[i];
            if (k == 0) {
                if (a != b) {
                    return -1;
                }
                continue;
            }
            if ((a - b) % k != 0) {
                return -1;
            }
            int y = (a - b) / k;
            ans += Math.abs(y);
            x += y;
        }
        return x == 0 ? ans / 2 : -1;
    }
}

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class Solution {
public:
    long long minOperations(vector<int>& nums1, vector<int>& nums2, int k) {
        long long ans = 0, x = 0;
        for (int i = 0; i < nums1.size(); ++i) {
            int a = nums1[i], b = nums2[i];
            if (k == 0) {
                if (a != b) {
                    return -1;
                }
                continue;
            }
            if ((a - b) % k != 0) {
                return -1;
            }
            int y = (a - b) / k;
            ans += abs(y);
            x += y;
        }
        return x == 0 ? ans / 2 : -1;
    }
};

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func minOperations(nums1 []int, nums2 []int, k int) int64 {
    ans, x := 0, 0
    for i, a := range nums1 {
        b := nums2[i]
        if k == 0 {
            if a != b {
                return -1
            }
            continue
        }
        if (a-b)%k != 0 {
            return -1
        }
        y := (a - b) / k
        ans += abs(y)
        x += y
    }
    if x != 0 {
        return -1