Array Partition I

Problem description

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example 1:

Input: [1,4,3,2]

Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

1. n is a positive integer, which is in the range of [1, 10000].

2. All the integers in the array will be in the range of [-10000, 10000].

Solution

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

Rough proof

For any optimal solution, sort the pairs by the smaller number of each pair in ascending order. Suppose the sorted pairs are $(a_1, b_1), ..., (a_n, b_n)$. Without loss of generality, assume $a_i \le b_i$.

We claim that $b_i \le a_{i+1}$ .

Suppose there is an optimal solution that has a group of two pairs $(a_i, b_i)$ and $(a_{i+1}, b_{i+1})$, such that $b_i > a_{i+1}$. We can re-group the four numbers to increase the total sum, i.e., $(a_i, a_{i+1}), (b_i, b_{i+1})$. A contradiction.