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

Solution 1: Mathematics

Let the minimum value of the array $\textit{nums}$ be $\textit{mi}$, the sum of the array be $\textit{s}$, and the length of the array be $\textit{n}$.

Assume the minimum number of operations is $\textit{k}$, and the final value of all elements in the array is $\textit{x}$. Then we have:

$$ \begin{aligned} \textit{s} + (\textit{n} - 1) \times \textit{k} &= \textit{n} \times \textit{x} \ \textit{x} &= \textit{mi} + \textit{k} \ \end{aligned} $$

Substituting the second equation into the first equation, we get:

$$ \begin{aligned} \textit{s} + (\textit{n} - 1) \times \textit{k} &= \textit{n} \times (\textit{mi} + \textit{k}) \ \textit{s} + (\textit{n} - 1) \times \textit{k} &= \textit{n} \times \textit{mi} + \textit{n} \times \textit{k} \ \textit{k} &= \textit{s} - \textit{n} \times \textit{mi} \ \end{aligned} $$

Therefore, the minimum number of operations is $\textit{s} - \textit{n} \times \textit{mi}$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array.

Python3

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

Java

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

C++

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();
    }
};

Go

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)
}

TypeScript

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;
}