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

Python3JavaC++GoTypeScript

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

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

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();
    }
};

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)
}

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;
    }
}

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