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Abhishek Chaudhary

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# Array Partition I

Given an integer array `nums` of `2n` integers, group these integers into `n` pairs `(a1, b1), (a2, b2), ..., (an, bn)` such that the sum of `min(ai, bi)` for all `i` is maximized. Return the maximized sum.

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:

1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4 So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Constraints:

• `1 <= n <= 104`
• `nums.length == 2 * n`
• `-104 <= nums[i] <= 104`

SOLUTION:

``````class Solution:
def arrayPairSum(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
total = 0
for i in range(0, n, 2):
total += min(nums[i:i + 2])