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.

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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) -&gt; min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -&gt; min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -&gt; 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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Constraints:


	1 &lt;= n &lt;= 104
	nums.length == 2 * n
	-104 &lt;= nums[i] &lt;= 104


class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        
        nums = sorted(nums)
        
        summ = 0
        for i in range(0,len(nums),2):
            summ += min(nums[i],nums[i+1])
        return summ
class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int sum = 0;
        for(int i = 0; i < nums.length; i+=2){
            sum += nums[i];
        }
        return sum;
    }
}
class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
     int res=0;
        sort(nums.begin(),nums.end());
        for(int i=0;i<nums.size();i+=2){
            res+=min(nums[i],nums[i+1]);  
        }
        return res;
    }
};
var arrayPairSum = function(nums) {
    nums.sort((a, b) => a - b);
    let total = 0;
    for (let i = 0; i < nums.length; i += 2) {
        total += Math.min(nums[i], nums[i + 1]);
    }
    return total;
};
Array Partition