Array Partition

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.
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// tc=O(nlogn)+O(n/2) sc=O(1)
class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        int n=nums.size();
        sort(nums.begin(), nums.end());
        int sum=0;
        for(int i=0;i<n;i=i+2){
           sum=sum+(min(nums[i], nums[i+1])) ;
        }
        return sum;
    }
};