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### by Sumit Chourasia | Oct 01, 2020 | Category :coding | Tags : algorithmडेटा-संरचनाआसानleetcodeगणित

#### Minimum Moves to Equal Array Elements - Math - Easy - LeetCode - मिनी टीवी

Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.

Example:

Input:
[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]

``````public class Solution {
public int MinMoves(int[] nums) {
if(nums.Length < 2){
return 0;
}
int min=Int32.MaxValue;
int sum=0;
foreach(int item in nums)
{
sum+=item;
min=Math.Min(min,item);
}

return sum-min*nums.Length;
}
}``````

Time Complexity: O(n)

Space Complexity: O(1)

let's define sum as the sum of all the numbers, before any moves; minNum as the min number int the list; n is the length of the list;

After say m moves, we get all the numbers as x, and we will get the following equation

sum + m * (n - 1) = x * n
and actually,

x = minNum + m
This part may be a little confusing, but @shijungg explained very well. let me explain a little again. it comes from two observations:

the minimum number will always be minimum until it reaches the final number, because every move, other numbers (besides the max) will be incremented too;
from above, we can get, the minimum number will be incremented in every move. So, if the final number is x, it would be minNum + moves;
and finally, we will get

sum - minNum * n = m
This is just a math calculation.

Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia