Generate 10-digit number using a phone keypad

Question

Given a phone keypad as shown below:

1 2 3
4 5 6
7 8 9
  0

How many different 10-digit numbers can be formed starting from 1? The constrain

Answer 1

I implemented both brute force and dynamic programming models

import queue


def chess_numbers_bf(start, length):
    if length <= 0:
        return 0
    phone = [[7, 5], [6, 8], [3, 7], [9, 2, 8], [], [6, 9, 0], [1, 5], [0, 2], [3, 1], [5, 3]]
    total = 0
    q = queue.Queue()
    q.put((start, 1))

    while not q.empty():
        front = q.get()
        val = front[0]
        len_ = front[1]
        if len_ < length:
            for elm in phone[val]:
                q.put((elm, len_ + 1))
        else:
            total += 1
    return total


def chess_numbers_dp(start, length):
    if length <= 0:
        return 0

    phone = [[7, 5], [6, 8], [3, 7], [9, 2, 8], [], [6, 9, 0], [1, 5], [0, 2], [3, 1], [5, 3]]
    memory = {}

    def __chess_numbers_dp(s, l):
        if (s, l) in memory:
            return memory[(s, l)]
        elif l == length - 1:
            memory[(s, l)] = 1
            return 1
        else:
            total_n_ways = 0
            for number in phone[s]:
                total_n_ways += __chess_numbers_dp(number, l+1)
            memory[(s, l)] = total_n_ways
            return total_n_ways
    return __chess_numbers_dp(start, 0)


# bf
for i in range(0, 10):
    print(i, chess_numbers_bf(3, i))
print('\n')

for i in range(0, 10):
    print(i, chess_numbers_bf(9, i))
print('\n')

# dp
for i in range(0, 10):
    print(i, chess_numbers_dp(3, i))
print('\n')

# dp
for i in range(0, 10):
    print(i, chess_numbers_dp(9, i))
print('\n')