# Solving Leetcode Interviews in Seconds with AI: Array Partition


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "561" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution to the problem:

*   **Sorting:** Sort the input array `nums` in ascending order.
*   **Pairing:** Iterate through the sorted array, taking pairs of adjacent elements.
*   **Summation:** Sum the minimum value of each pair. Because the array is sorted, the minimum value of each pair will always be the first element of the pair.

*   **Runtime Complexity:** O(n log n) due to sorting. **Storage Complexity:** O(1) or O(n), depending on the sorting algorithm's space complexity (in-place sorting algorithms can achieve O(1), while others might use O(n)).

	
	# Code
	```python
	def array_pair_sum(nums):
    """
    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.

    Args:
    nums (List[int]): An array of integers.

    Returns:
    int: The maximized sum of the minimum values in each pair.
    """
    nums.sort()
    max_sum = 0
    for i in range(0, len(nums), 2):
        max_sum += nums[i]
    return max_sum
	```