Question

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

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

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;
  }
}