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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