Step Number from 10 to N using python

Question

The Program must accept an integer N. The program must print all the stepping number from 10 to N, if there is no such number present the program should print -1 as

Answer 1

Maybe the main trick here is that the max range is 10^7. If we consider each digit as a node of the graph, we can traverse it with bfs/dfs and at each point, we can only move to the adjacent node (digit + 1, digit -1). Because the maximum depth is only 7, the solution should be pretty fast.

Here's a rough DFS implementation, you can improve the details.

sol_list = []
def dfs(i, num, N):
  # print(i, num)
  if num > N: # too much, need to break
    return
  if num <= N and num >= 10: # perfect range, add to solution, I can add some repeated solution as I called dfs(i+1,0,N) multiple times
    global sol_list
    # print(num)
    sol_list.append(num) # add to solution

  if i > 7:
    return
  if num == 0: # I can call another 0
    dfs(i+1, 0, N) # this is not needed if the numbers are added in the previous list without checking
  last_dig = num % 10
  if last_dig == 0:
      dfs(i+1, num*10 + 1, N) # if last digit is 0, we can only choose 1 as next digit not -1
  elif last_dig == 9:
    dfs(i+1, num*10 + 8, N)
  else:
    dfs(i+1, num*10 + (last_dig-1), N)
    dfs(i+1, num*10 + (last_dig+1), N)

import time
t1 = time.time()
[dfs(0, i, 10000000) for i in range(10)] # staring with every digit possible 0-9
t2 = time.time()
print(t2-t1)
sol = sorted(set(sol_list))
print(sol) # added some repeated solutions, it's easier to set them