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# Array Partition LeetCode Solution

## Problem – 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.``````

Constraints:

• `1 <= n <= 104`
• `nums.length == 2 * n`
• `-104 <= nums[i] <= 104`

### Array Partition LeetCode Solution in Java

``````public class Solution {
public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int result = 0;
for (int i = 0; i < nums.length; i += 2) {
result += nums[i];
}
return result;
}
}
``````

### Array Partition LeetCode Solution in Python

``````class Solution(object):

def arrayPairSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return sum(sorted(nums)[::2])
``````

### Array Partition LeetCode Solution in C++

``````class Solution {
public:
int arrayPairSum(vector<int>& nums) {
vector<int> hashtable(20001,0);
for(size_t i=0;i<nums.size();i++)
{
hashtable[nums[i]+10000]++;
}
int ret=0;
int flag=0;
for(size_t i=0;i<20001;){
if((hashtable[i]>0)&&(flag==0)){
ret=ret+i-10000;
flag=1;
hashtable[i]--;
}else if((hashtable[i]>0)&&(flag==1)){
hashtable[i]--;
flag=0;
}else i++;
}
return ret;
}
};
``````
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