453. Minimum Moves to Equal Array Elements

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]

Approach #1: Math. [Java]

class Solution {
    public int minMoves(int[] nums) {
        int len = nums.length;
        int sum = 0, minNum = Integer.MAX_VALUE;
        for (int n : nums) {
            sum += n;
            minNum = Math.min(minNum, n);
        }
        return sum - minNum * len;
    }
}

　　

Analysis:

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;

It comes from two observations:

The minum number will always be minum unitl it reachs the final number, because every move, other numbers (besides the max) will be increamented too;

From above, we can get, the minum 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.

Reference:

https://leetcode.com/problems/minimum-moves-to-equal-array-elements/discuss/93817/It-is-a-math-question