Maximize Score After N Operations - DP w/ Bitmasks [JS]

Description 

Solution: DP w/ Bitmasks

Memoize each dp(mask), where mask = bitmask of which numbers we have removed.
Since we know how many operations we have done based on the bitmask, we don't need to memoize i.

For the ith operation, try every pair of unused numbers and record the maximum score.

m = length of nums
Time Complexity: O(2^m * m^2) 420ms
Space Complexity: O(2^m) 44.5MB

var maxScore = function(nums) {
  let n = nums.length / 2, m = nums.length;
  let memo = Array(1 << m).fill(-1);
  return dp(1, 0);
  
  function dp(i, mask) {
    if (i === n + 1) return 0;
    if (memo[mask] !== -1) return memo[mask];
    
    let ans = 0;
    for (let j = 0; j < m; j++) {
      if ((mask >> j) & 1) continue; // j has already been used
      for (let k = j + 1; k < m; k++) {
        if ((mask >> k) & 1) continue; // k has already been used
        let score = i * gcd(nums[j], nums[k]);
        let newMask = mask | (1 << j) | (1 << k);
        ans = Math.max(ans, dp(i + 1, newMask) + score);
      }
    }
    return memo[mask] = ans;
  }
  
  function gcd(a, b) {
    if (b === 0) return a;
    return gcd(b, a % b);
  }
};