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.

• Java
• C++
• Python
• Go
• TypeScript
• C#
• Javascript
• RenderScript

• 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;
      }
  }
  
  

• /**
   * @param {number[]} nums
   * @return {number}
   */
  var minPairSum = function (nums) {
      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;
  };
  
  

• impl Solution {
      pub fn min_pair_sum(nums: Vec<i32>) -> i32 {
          let mut nums = nums;
          nums.sort();
          let mut ans = 0;
          let n = nums.len();
          for i in 0..n / 2 {
              ans = ans.max(nums[i] + nums[n - i - 1]);
          }
          ans
      }
  }
  
  

All Problems

All Solutions