Array Partition

Description

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

Code

Time Complexity: $O(N\log N)$, Space Complexity: $O(1)$

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
       sort(nums.begin(), nums.end());
       int res = 0;
       for(int i = 0; i < nums.size(); i += 2) {
           res += nums[i];
       } 
       return res;
    }
};

Source

• Array Partition - LeetCode