Switch indexes, create the next highest number using the same exact 9 digits presented

Question

So I received a challenge that states the following: \"Design a program that takes as input a 9 digit number where no digit appears twice and produces as output an arrangeme

Answer 1

I am not convinced that your approach of flipping digits is guaranteed to find the next highest number (at least not without further checks)

Here a simple solution: Simply increment the input number and check if the conditions are met or if no number can be found.

set() can be used to get the set of unique digits in the number.

input_num = '781623954'
next_num = int(input_num) + 1
input_digits = set(input_num)
found = False
while not found:
    next_num += 1
    next_digits = set(str(next_num))
    found = len(next_digits) == 9 and input_digits == next_digits
    if next_num > 987654321:
        break

if found:
    print(next_num)
else:
    print("No number was found.")