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Leetcode 2815. Max Pair Sum in an Array

Problem Statement

The problem is given as: You are given a 0-indexed integer array nums. You can pair two numbers in the array and take the sum of each pair. Find the pair with the maximum sum and return that sum.

Example:

Input: nums = [5, 9, 7, 3]
Output: 16
Explanation: The pairs can be (5+9), (5+7), (5+3), (9+7), (9+3), (7+3). The maximum sum is 16.

Clarifying Questions

  1. Q: Does the array contain negative numbers? A: Yes, the array can contain any integers, including negative numbers.

  2. Q: How large can the array be? A: The array size can be reasonably large, but it should fit within typical constraints for array size in competitive programming (up to (10^5) elements, usually).

  3. Q: Can the array be empty? A: We should assume the array has at least two elements, as pairing requires at least two numbers.

  4. Q: Should we consider the same element twice for a pair? A: No, each element can only be paired with another distinct element.

Strategy

To solve this problem efficiently:

  1. Sorting Approach:
    • We can sort the array.
    • The maximum pair sum would be the sum of the two largest elements.
  2. This would yield the correct result because the sum of any other pair would be less than or equal to the sum of the two largest elements after the array is sorted.

Solution and Code

Here’s a potential implementation in C++:

#include <vector>
#include <algorithm>
#include <iostream>

int maxPairSum(std::vector<int>& nums) {
    std::sort(nums.begin(), nums.end());
    int n = nums.size();
    return nums[n-1] + nums[n-2]; // Sum of the two largest elements
}

int main() {
    std::vector<int> nums = {5, 9, 7, 3};
    std::cout << "Max Pair Sum: " << maxPairSum(nums) << std::endl; // Output should be 16
    return 0;
}

Time Complexity

Additional Notes

If the array is already sorted or if there are constraints that ensure the array is in a particular order, this solution could be adjusted for optimization. For instance, if we can guarantee a fast look-up for the two largest elements, a linear-time solution might be appropriate.

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