# Solving Leetcode Interviews in Seconds with AI: Maximize Score After N Operations


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1799" 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
	> You are given nums, an array of positive integers of size 2 * n. You must perform n operations on this array. In the ith operation (1-indexed), you will:  Choose two elements, x and y. Receive a score of i * gcd(x, y). Remove x and y from nums.  Return the maximum score you can receive after performing n operations. The function gcd(x, y) is the greatest common divisor of x and y.   Example 1:  Input: nums = [1,2] Output: 1 Explanation: The optimal choice of operations is: (1 * gcd(1, 2)) = 1  Example 2:  Input: nums = [3,4,6,8] Output: 11 Explanation: The optimal choice of operations is: (1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11  Example 3:  Input: nums = [1,2,3,4,5,6] Output: 14 Explanation: The optimal choice of operations is: (1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14    Constraints:  1 <= n <= 7 nums.length == 2 * n 1 <= nums[i] <= 106  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming with Bitmasking:** Represent the state of `nums` using a bitmask, where each bit indicates whether a number has been removed.  Use dynamic programming to explore all possible pairings and store the maximum score achievable for each state.
*   **GCD Optimization:** Precompute the GCD of all possible pairs of numbers in `nums` to avoid redundant calculations within the dynamic programming loop.
*   **Memoization:** Store the results of subproblems (maximum score for a given state) to avoid recomputation, significantly improving efficiency.

*   **Runtime Complexity: O(n<sup>2</sup> * 2<sup>2n</sup>), Storage Complexity: O(2<sup>2n</sup> + n<sup>2</sup>)**

	
	# Code
	```python
	import math

def max_score(nums):
    n = len(nums) // 2
    memo = {}
    gcd_matrix = {}

    def gcd(a, b):
        if (a, b) in gcd_matrix:
            return gcd_matrix[(a, b)]
        if b == 0:
            gcd_matrix[(a, b)] = a
            gcd_matrix[(b, a)] = a
            return a
        gcd_matrix[(a, b)] = gcd(b, a % b)
        gcd_matrix[(b, a)] = gcd_matrix[(a, b)]
        return gcd_matrix[(a, b)]

    def solve(mask, op):
        if op > n:
            return 0
        if (mask, op) in memo:
            return memo[(mask, op)]

        max_score_val = 0
        for i in range(2 * n):
            if (mask >> i) & 1 == 0:
                for j in range(i + 1, 2 * n):
                    if (mask >> j) & 1 == 0:
                        new_mask = mask | (1 << i) | (1 << j)
                        current_score = op * gcd(nums[i], nums[j]) + solve(new_mask, op + 1)
                        max_score_val = max(max_score_val, current_score)
        memo[(mask, op)] = max_score_val
        return max_score_val

    return solve(0, 1)
	```