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Abhishek Chaudhary

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# Minimize Maximum Pair Sum in Array

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`

SOLUTION:

``````import heapq

class Solution:
def minPairSum(self, nums: List[int]) -> int:
minheap = []
maxheap = []
for num in nums:
heapq.heappush(minheap, num)
heapq.heappush(maxheap, -num)
n = len(nums)
smallest = heapq.nlargest(n // 2, minheap)
biggest = heapq.nlargest(n // 2, maxheap)
return max(smallest[i] - biggest[i] for i in range(n // 2))
``````