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.