Question

1877. Minimize Maximum Pair Sum in Array

中文文档

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

Solution 1

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]))

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;
  }
};

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;
  }
}