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561. Array Partition

Description

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

 

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

 

Constraints:

  • 1 <= n <= 104
  • nums.length == 2 * n
  • -104 <= nums[i] <= 104

Solutions

  • class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int ans = 0;
        for (int i = 0; i < nums.length; i += 2) {
            ans += nums[i];
        }
        return ans;
    }
}

  • class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 0;
        for (int i = 0; i < nums.size(); i += 2) ans += nums[i];
        return ans;
    }
};

  • class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        return sum(sorted(nums)[::2])


  • func arrayPairSum(nums []int) int {
	sort.Ints(nums)
	ans := 0
	for i := 0; i < len(nums); i += 2 {
		ans += nums[i]
	}
	return ans
}

  • /**
 * @param {number[]} nums
 * @return {number}
 */
var arrayPairSum = function (nums) {
    nums.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0; i < nums.length; i += 2) {
        ans += nums[i];
    }
    return ans;
};


  • impl Solution {
    pub fn array_pair_sum(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let n = nums.len();
        let mut i = 0;
        let mut res = 0;
        while i < n {
            res += nums[i];
            i += 2;
        }
        res
    }
}

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