Maximize Score After N Operations

Problem

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:&nbsp;The optimal choice of operations is:
(1 * gcd(1, 2)) = 1

Example 2:

Input: nums = [3,4,6,8]
Output: 11
Explanation:&nbsp;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:&nbsp;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] <= 10^6

Solution

class Solution {
    public int maxScore(int[] nums) {
        int n = nums.length;
        Integer[] memo = new Integer[1 << n];
        return helper(1, 0, nums, memo);
    }

    private int helper(int operationNumber, int mask, int[] nums, Integer[] memo) {
        int n = nums.length;
        if (memo[mask] != null) {
            return memo[mask];
        }
        if (operationNumber > n / 2) {
            return 0;
        }
        int maxScore = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            if ((mask & (1 << i)) == 0) {
                for (int j = i + 1; j < n; j++) {
                    if ((mask & (1 << j)) == 0) {
                        int score = operationNumber * gcd(nums[i], nums[j]);
                        int score2 =
                                helper(operationNumber + 1, mask | (1 << i) | (1 << j), nums, memo);
                        maxScore = Math.max(maxScore, score + score2);
                    }
                }
            }
        }
        memo[mask] = maxScore;
        return maxScore;
    }

    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).