# 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.

• 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;
}
}

• 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;
}
};

• 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


• 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
}
return int64(ans / 2)
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function minOperations(nums1: number[], nums2: number[], k: number): number {
const n = nums1.length;
if (k === 0) {
return nums1.every((v, i) => v === nums2[i]) ? 0 : -1;
}
let sum1 = 0;
let sum2 = 0;
for (let i = 0; i < n; i++) {
const diff = nums1[i] - nums2[i];
sum1 += diff;
if (diff % k !== 0) {
return -1;
}
sum2 += Math.abs(diff);
}
if (sum1 !== 0) {
return -1;
}
return sum2 / (k * 2);
}


• impl Solution {
pub fn min_operations(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> i64 {
let k = k as i64;
let n = nums1.len();
if k == 0 {
return if
nums1
.iter()
.enumerate()
.all(|(i, &v)| v == nums2[i])
{
0
} else {
-1
};
}
let mut sum1 = 0;
let mut sum2 = 0;
for i in 0..n {
let diff = (nums1[i] - nums2[i]) as i64;
sum1 += diff;
if diff % k != 0 {
return -1;
}
sum2 += diff.abs();
}
if sum1 != 0 {
return -1;
}
sum2 / (k * 2)
}
}