Array Partition I

561. Array Partition I

 Given an integer array nums of 2n integers, group these integers into n pairs (a₁, b₁), (a₂, b₂), ..., (a_n, b_n) such that the sum of min(a_i, b_i) 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 to the above Question

class Solution {

public:

    int arrayPairSum(vector<int>& nums) {

        sort(nums.begin(), nums.end());

        int a=0;

        for(int i=0; i<nums.size(); i+=2){

            a=a+nums[i];

        }

        return a;

    }

};