# 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

```cpp
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.


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