#1877

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Minimize Maximum Pair Sum in Array

Minimize the maximum pair sum in an array by optimally pairing its elements.

Problem Statement

Given an array of even length, you need to pair up its elements optimally to minimize the maximum pair sum. The pair sum of two elements is the sum of the pair (a, b) where sum = a + b. Your task is to return the minimized maximum sum of all pairs formed.

You should consider using sorting to help with pairing the elements. By pairing the smallest and largest elements first, you can balance the sum and reduce the maximum pair sum.

Examples

Example 1

Input: nums = [3,5,2,3]

Output: 7

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

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

Solution Approach

Sort and Two-pointer Approach

Sort the array and use two pointers: one starting from the beginning and the other from the end. Pair the elements at these pointers together, and adjust the pointers inward, ensuring that you always pair the smallest unpaired element with the largest.

Greedy Pairing Strategy

The greedy approach minimizes the maximum pair sum by ensuring that the sum of each pair is as balanced as possible. By sorting the array and pairing the extremes, this strategy aims to reduce the worst-case pair sum.

Invariant Tracking

Maintain an invariant during the two-pointer scanning process, tracking the maximum pair sum as you pair the elements. This invariant ensures that you always have the optimal configuration without needing to reevaluate the entire array.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Sorting the array takes O(n log n) time. The two-pointer scanning process runs in O(n), so the overall time complexity is O(n log n). Space complexity is O(1) if the sorting is done in-place.

What Interviewers Usually Probe

Look for understanding of the greedy pairing technique.
Evaluate the candidate's ability to optimize through sorting.
Check if the candidate efficiently handles larger arrays with the two-pointer approach.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array before pairing, leading to suboptimal pairings.
Misunderstanding the importance of balancing the pairs to minimize the maximum sum.
Overcomplicating the problem by attempting to pair elements randomly rather than using a structured approach.

Follow-up variants

Pairing elements with constraints on pair sums.
Optimizing pair sums for a larger range of values.
Extending the problem to odd-length arrays with different pairing strategies.

FAQ

How do you minimize the maximum pair sum in the array?

The optimal approach is to sort the array and then use two-pointer scanning to pair the smallest and largest elements, ensuring balanced pair sums.

What is the time complexity of the solution?

The time complexity is O(n log n) due to the sorting step, and the two-pointer scanning takes O(n) time.

Can the solution be optimized further?

The solution is already quite efficient with O(n log n) complexity. Further optimization is unlikely unless the problem constraints change.

Why does sorting help in this problem?

Sorting allows you to pair the smallest and largest elements, minimizing the maximum sum and achieving a more balanced solution.

What pattern is used in this problem?

This problem primarily uses a two-pointer scanning technique combined with a greedy strategy and sorting to achieve optimal pairings.

terminal

Solution

Solution 1: Greedy

To minimize the maximum pair sum in the array, we can pair the smallest number with the largest number, the second smallest with the second largest, and so on.

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

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