# 453. Minimum Moves to Equal Array Elements

## Description

Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.

In one move, you can increment n - 1 elements of the array by 1.

Example 1:

Input: nums = [1,2,3]
Output: 3
Explanation: Only three moves are needed (remember each move increments two elements):
[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]


Example 2:

Input: nums = [1,1,1]
Output: 0


Constraints:

• n == nums.length
• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109
• The answer is guaranteed to fit in a 32-bit integer.

## Solutions

• class Solution {
public int minMoves(int[] nums) {
return Arrays.stream(nums).sum() - Arrays.stream(nums).min().getAsInt() * nums.length;
}
}


• class Solution {
public:
int minMoves(vector<int>& nums) {
int s = 0;
int mi = 1 << 30;
for (int x : nums) {
s += x;
mi = min(mi, x);
}
return s - mi * nums.size();
}
};

• class Solution:
def minMoves(self, nums: List[int]) -> int:
return sum(nums) - min(nums) * len(nums)


• func minMoves(nums []int) int {
mi := 1 << 30
s := 0
for _, x := range nums {
s += x
if x < mi {
mi = x
}
}
return s - mi*len(nums)
}

• function minMoves(nums: number[]): number {
let mi = 1 << 30;
let s = 0;
for (const x of nums) {
s += x;
mi = Math.min(mi, x);
}
return s - mi * nums.length;
}