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.

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Example 1:

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


Example 2:

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


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Constraints:


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


class Solution:
    def minMoves(self, nums: List[int]) -> int:
        return sum(nums)-min(nums)*len(nums)
class Solution {
    public int minMoves(int[] nums) {
        int min=Integer.MAX_VALUE;
        int count=0;
        for(int i:nums)
            min=Math.min(i,min);
        
        for(int i=0;i<nums.length;i++)
        {
            count+=nums[i]-min;
        }
        return count;
    }
}
class Solution {
public:
    int minMoves(vector<int>& nums) {

        // sorting the array to get min at the first
        sort(nums.begin(), nums.end());
        int cnt = 0, n = nums.size();

        // Now we have to make min equal to every number and keep adding the count
        for(int i = 1; i < n; i++)
            cnt += nums[i] - nums[0];

        return cnt;
    }
};
var minMoves = function(nums) {
    let minElm = Math.min(...nums);
    let ans = 0;
    for(let i=0; i<nums.length; i++){
        ans += (nums[i]-minElm)
    }
    return ans
};
Minimum Moves to Equal Array Elements