Welcome to Subscribe On Youtube

1877. Minimize Maximum Pair Sum in Array

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.

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

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

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


  • func minPairSum(nums []int) (ans int) {
	sort.Ints(nums)
	n := len(nums)
	for i, x := range nums[:n>>1] {
		ans = max(ans, x+nums[n-1-i])
	}
	return
}

  • function minPairSum(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < n >> 1; ++i) {
        ans = Math.max(ans, nums[i] + nums[n - 1 - i]);
    }
    return ans;
}


  • public class Solution {
    public int MinPairSum(int[] nums) {
        Array.Sort(nums);
        int ans = 0, n = nums.Length;
        for (int i = 0; i < n >> 1; ++i) {
            ans = Math.Max(ans, nums[i] + nums[n - i - 1]);
        }
        return ans;
    }
}

All Problems

All Solutions