Hamiltonian Flights - Implementation

function hamiltonianFlights(n, edges):
    adj = adjacency list
    dp = 2D array of size [2^n][n], initialized to 0
    MOD = 10^9 + 7

    dp[1][0] = 1 // Start at city 1, only city 1 visited

    for mask from 1 to 2^n - 1:
        for last from 0 to n-1:
            if not (mask & (1 << last)):
                continue
            if dp[mask][last] == 0:
                continue
            for next in adj[last]:
                if mask & (1 << next):
                    continue
                newMask = mask | (1 << next)
                dp[newMask][next] = (dp[newMask][next] + dp[mask][last]) % MOD

    return dp[(1 << n) - 1][n-1]

Time: $O(n^2 \cdot 2^n)$. With $n = 20$, this is about $4 \times 10^8$ operations.

Space complexity is $O(n \cdot 2^n)$ for the data structures used.