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462. Minimum Moves to Equal Array Elements II

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 or decrement an element of the array by 1.

Test cases are designed so that the answer will fit in a 32-bit integer.

 

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation:
Only two moves are needed (remember each move increments or decrements one element):
[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

Example 2:

Input: nums = [1,10,2,9]
Output: 16

 

Constraints:

  • n == nums.length
  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Solutions

  • class Solution {
    public int minMoves2(int[] nums) {
        Arrays.sort(nums);
        int k = nums[nums.length >> 1];
        int ans = 0;
        for (int v : nums) {
            ans += Math.abs(v - k);
        }
        return ans;
    }
}

  • class Solution {
public:
    int minMoves2(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int k = nums[nums.size() >> 1];
        int ans = 0;
        for (int& v : nums) ans += abs(v - k);
        return ans;
    }
};

  • class Solution:
    def minMoves2(self, nums: List[int]) -> int:
        nums.sort()
        k = nums[len(nums) >> 1]
        return sum(abs(v - k) for v in nums)


  • func minMoves2(nums []int) int {
	sort.Ints(nums)
	k := nums[len(nums)>>1]
	ans := 0
	for _, v := range nums {
		ans += abs(v - k)
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

  • function minMoves2(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const mid = nums[nums.length >> 1];
    return nums.reduce((r, v) => r + Math.abs(v - mid), 0);
}


  • impl Solution {
    pub fn min_moves2(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let mid = nums[nums.len() / 2];
        let mut res = 0;
        for num in nums.iter() {
            res += (num - mid).abs();
        }
        res
    }
}

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