README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0453.Minimum%20Moves%20to%20Equal%20Array%20Elements/README_EN.md	Array Math

453. Minimum Moves to Equal Array Elements

中文文档

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

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;
}

Solution 2

Java

class Solution {
    public int minMoves(int[] nums) {
        int s = 0;
        int mi = 1 << 30;
        for (int x : nums) {
            s += x;
            mi = Math.min(mi, x);
        }
        return s - mi * nums.length;
    }
}