Welcome to a detailed guide where we’ll unravel the intricacies of optimizing pair sums in arrays, as we dive into solving the intriguing LeetCode problem 1877: ‘Minimize Maximum Pair Sum in Array’.

Introduction

In this tutorial, we’ll tackle the LeetCode problem “1877. Minimize Maximum Pair Sum in Array.” This problem is a fascinating blend of array manipulation and optimization, suitable for learners who are comfortable with basic programming concepts.

Problem Statement

Given an even-length array nums, our goal is to pair up elements in such a way that the maximum pair sum is as small as possible. The pair sum of two elements (a, b) is a + b. We need to find an optimal pairing strategy that minimizes the largest pair sum.

Examples

1. Example 1:

• Input: nums = [3,5,2,3]
• Output: 7
• Explanation: Pair the elements as (3,3) and (5,2). The maximum pair sum is max(6, 7) = 7.

1. Example 2:

• Input: nums = [3,5,4,2,4,6]
• Output: 8
• Explanation: Pair the elements as (3,5), (4,4), and (6,2). The maximum pair sum is max(8, 8, 8) = 8.

Solution Approach

Our approach involves sorting the array and then pairing the smallest and largest elements together. This strategy ensures that the sum of these pairs is as small as possible.

Step-by-Step Solution

Sort the Array:

• First, sort the array in ascending order. This helps us easily pair the smallest and largest elements.

Initialize Pointers:

• Use two pointers, left starting at the beginning of the array and right at the end.

Find Minimized Maximum Pair Sum:

• Iterate through the array, pairing elements pointed by left and right.
• Calculate the pair sum and update the maximum pair sum accordingly.
• Move left forward and right backward until they meet.

Return the Result:

• The minimized maximum pair sum is the result.

Implementation

Here’s the Python code implementing the above logic:

class Solution:
    def minPairSum(self, nums: List[int]) -> int:
        # Sort the array to facilitate pairing of smallest and largest elements
        nums.sort()

        # Initialize two pointers: 'left' at the start and 'right' at the end of the array
        left, right = 0, len(nums) - 1

        # Initialize min_max_pair_sum to store the maximum pair sum found so far
        # We start with negative infinity as we are looking for the maximum
        min_max_pair_sum = float('-inf')

        # Loop through the array until the two pointers meet
        while left < right:
            # Calculate the sum of the pair pointed by 'left' and 'right'
            current_pair_sum = nums[left] + nums[right]

            # Update min_max_pair_sum with the maximum of the current pair sum and the previous min_max_pair_sum
            min_max_pair_sum = max(min_max_pair_sum, current_pair_sum)

            # Move 'left' pointer forward and 'right' pointer backward to form the next pair
            left += 1
            right -= 1
        
        # Return the maximum pair sum after considering all pairs
        return min_max_pair_sum

Complexity Analysis

• Time Complexity: O(n log n) due to sorting.
• Space Complexity: O(1) as we use constant extra space.

Conclusion

This problem demonstrates the effectiveness of sorting in optimizing pairings within an array. By carefully pairing elements, we can minimize the maximum pair sum, showcasing a clever use of two-pointer technique in array manipulation.