# LeetCode 1799. Maximize Score After N Operations

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] <= 10^6`

## Solution

[English Version in Youtube](https://youtu.be/iYUoydllPc4)

[中文版解答Youtube Link](https://youtu.be/94vDosERuiI)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1g64y1D7j6/)

```
class Solution {
    int dfs(const vector<int>& nums, int round, int state, unordered_map<int, int>& cache) {
        if (round > nums.size() / 2) {
            return 0;
        }
        if (cache.count(state) > 0) {
            return cache[state];
        }
        
        int ans = 0;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                int picked = (1 << i) | (1 << j);
                if ((state & picked) == 0) {
                    ans = max(ans, round * __gcd(nums[i], nums[j]) + dfs(nums, round + 1, state | picked, cache));
                }
            }
        }

        return cache[state] = ans;
    }
public:
    int maxScore(vector<int>& nums) {
        unordered_map<int, int> cache;
        return dfs(nums, /*round=*/1, /*state=*/0, cache);
    }
};
```


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