maximum AND sum of array

You are given an integer array nums of length n and an integer numSlots such that 2 * numSlots >= n. There are numSlots slots numbered from 1 to numSlots.
You have to place all n integers into the slots such that each slot contains at most two numbers. The AND sum of a given placement is the sum of the bitwise AND of every number with its respective slot number.
For example, the AND sum of placing the numbers [1, 3] into slot 1 and [4, 6] into slot 2 is equal to (1 AND 1) + (3 AND 1) + (4 AND 2) + (6 AND 2) = 1 + 1 + 0 + 2 = 4.
Return the maximum possible AND sum of nums given numSlots slots.
* Example 1:
Input: nums = [1,2,3,4,5,6], numSlots = 3
Output: 9
Explanation: One possible placement is [1, 4] into slot 1, [2, 6] into slot 2, and [3, 5] into slot 3. 
This gives the maximum AND sum of (1 AND 1) + (4 AND 1) + (2 AND 2) + (6 AND 2) + (3 AND 3) + (5 AND 3) = 1 + 0 + 2 + 2 + 3 + 1 = 9.
* Example 2:
Input: nums = [1,3,10,4,7,1], numSlots = 9
Output: 24
Explanation: One possible placement is [1, 1] into slot 1, [3] into slot 3, [4] into slot 4, [7] into slot 7, and [10] into slot 9.
This gives the maximum AND sum of (1 AND 1) + (1 AND 1) + (3 AND 3) + (4 AND 4) + (7 AND 7) + (10 AND 9) = 1 + 1 + 3 + 4 + 7 + 8 = 24.
Note that slots 2, 5, 6, and 8 are empty which is permitted.
Constraints:
n == nums.length
1 <= numSlots <= 9
1 <= n <= 2 * numSlots
1 <= nums[i] <= 15

```
class Solution {
public:
    int hammingWeight(int n) {
        int cnt=0;  
        while(n>0){ 
            if((n&1)>0) 
                ++cnt;
            n=n>>1; 
        }
        return cnt;
    }
    
    int maximumANDSum(vector<int>& nums, int numSlots) {
        nums.insert(nums.end(), 2*numSlots-nums.size(), 0);
        
        int L = nums.size();
        int N = 1<<L;
        vector<int> results(N);
        for (int i = 1; i < N; i++) {
            int cnt = hammingWeight(i), slot = (cnt + 1) / 2; 
            for (int j = 0; j < L; j++) {
                if (i & (1<<j)) { 
                    results[i] = max(results[i], results[i ^ (1<<j)] + (slot & nums[j]));
                }
            }
        }
        return results[N-1];        
    }
};
```
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