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

You can simplify the generation of numbers by noting that each time a digit is added to an existing stepping number it must be either 1 more or 1 less than the existing last digit. So we can generate all the stepping numbers with a given number of digits by starting with single digit numbers (1-9) and then repeatedly adding digits to them until we have reached the number of digits we want. So for example, starting with the digit 1, and needing to go to 4 digits, we would produce

1 => 10, 12
10, 12 => 101, 121, 123
101, 121, 123 => 1010, 1012, 1210, 1212, 1232, 1234

The number of digits we need is computed by using math.ceil(math.log10(N)).

import math

def stepNums(N):
    if N < 10:
        return -1
    digits = math.ceil(math.log10(N))
    sn = [[]] * digits
    # 1 digit stepping numbers
    sn[0] = list(range(1, 10))
    # m digit stepping numbers
    for m in range(1, digits):
        sn[m] = []
        for s in sn[m-1]:
            if s % 10 != 0:
                sn[m].append(s * 10 + s % 10 - 1)
            if s % 10 != 9:
                sn[m].append(s * 10 + s % 10 + 1)
    return [s for l in sn for s in l if 10 <= s <= N]

e.g.

print(stepNums(3454))

Output:

[10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 101, 121, 123, 210, 212, 232, 234, 321, 323, 343, 345, 432, 434, 454, 456, 543, 545, 565, 567, 654, 656, 676, 678, 765, 767, 787, 789, 876, 878, 898, 987, 989, 1010, 1012, 1210, 1212, 1232, 1234, 2101, 2121, 2123, 2321, 2323, 2343, 2345, 3210, 3212, 3232, 3234, 3432, 3434, 3454]

Note that the code could potentially be sped up by comparing the generated numbers to N so that when calling stepNums(10001) we don't generate all the numbers up to 98989.