Minimum Cost to Connect Two Groups of Points - DP w/ Bitmasks [JS]

Description 

Solution: DP w/ Bitmasks

Memoize each dp(i, mask), where
i = index in group1
mask = bitmask of points connected in group2

First, connect every point in group1 with a point in group2. Try every combination and record the best result.
Then, connect every unconnected point in group2 with the points in group1 with the cheapest connection cost.

n = size of group1, m = size of group2
Time Complexity: O(n * 2^m * m) 161ms
Space Complexity: O(n * 2^m) 44.5MB

var connectTwoGroups = function(cost) {
  let n = cost.length, m = cost[0].length; // n = size of group1, m = size of group2
  let minCost = Array(m).fill(Infinity); // minCost[j] = minimum cost to connect point j (from group2) to a point in group1
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      minCost[j] = Math.min(minCost[j], cost[i][j]);
    }
  }
  let memo = Array(n).fill(0).map(() => Array(1 << m).fill(-1));
  return dp(0, 0);
  
  function dp(i, mask) {
    if (i === n) {
      let cost = 0;
      for (let j = 0; j < m; j++) {
        if (((mask >> j) & 1) === 0) cost += minCost[j];
      }
      return cost;
    }
    if (memo[i][mask] !== -1) return memo[i][mask];
    
    let ans = Infinity;
    for (let j = 0; j < m; j++) {
      ans = Math.min(ans, dp(i + 1, mask | (1 << j)) + cost[i][j]);
    }
    return memo[i][mask] = ans;
  }  
};