README_EN.md

1877. Minimize Maximum Pair Sum in Array

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Description

The pair sum of a pair (a,b) is equal to a + b. The maximum pair sum is the largest pair sum in a list of pairs.

• For example, if we have pairs (1,5), (2,3), and (4,4), the maximum pair sum would be max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8.

Given an array nums of even length n, pair up the elements of nums into n / 2 pairs such that:

• Each element of nums is in exactly one pair, and
• The maximum pair sum is minimized.

Return the minimized maximum pair sum after optimally pairing up the elements.

 

Example 1:

Input: nums = [3,5,2,3]

Output: 7

Explanation: The elements can be paired up into pairs (3,3) and (5,2).

The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.

Example 2:

Input: nums = [3,5,4,2,4,6]

Output: 8

Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).

The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.

 

Constraints:

• n == nums.length
• 2 <= n <= 105
• n is even.
• 1 <= nums[i] <= 105

Solutions

Sort & Greedy.

Python3

class Solution:
    def minPairSum(self, nums: List[int]) -> int:
        nums.sort()
        res, n = 0, len(nums)
        for i in range(n >> 1):
            res = max(res, nums[i] + nums[n - i - 1])
        return res

Java

class Solution {
    public int minPairSum(int[] nums) {
        Arrays.sort(nums);
        int res = 0, n = nums.length;
        for (int i = 0; i < (n >> 1); ++i) {
            res = Math.max(res, nums[i] + nums[n - i - 1]);
        }
        return res;
    }
}

C++

class Solution {
public:
    int minPairSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int res = 0, n = nums.size();
        for (int i = 0; i < (n >> 1); ++i) {
            res = max(res, nums[i] + nums[n - i - 1]);
        }
        return res;
    }
};

Go

func minPairSum(nums []int) int {
	sort.Ints(nums)
	res, n := 0, len(nums)
	for i := 0; i < (n >> 1); i++ {
		res = max(res, nums[i]+nums[n-i-1])
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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