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

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 arrangement of the same 9 digits corresponding to the next highest number. If no such number exists, the algorithm should indicate this. So for example, if the input is 781623954 the output would be 781624359."

So I came up with this idea to flip the indexes, so check the last index with the one right before to see which is bigger and compare then flip if necessary but for some reason my code isn't working. I only did the work for checking the last two digits not all the digits, so if you can help me out and check it for me and if you have any better ideas on how to tackle this problem, please share.

input = raw_input("Enter 9 Digits: ")
x = 9
while x>0:
    x-=1
    if input[8] > input[7]:
        temp = input[8]
        input[8] == input[7] 
        input[7] == temp
        print input
        break

Solution 1:

Here's a more efficient approach, using the algorithm of the 14th century Indian mathematician Narayana Pandita, which can be found in the Wikipedia article on Permutation. This ancient algorithm is still one of the fastest known ways to generate permutations in order, and it is quite robust, in that it properly handles permutations that contain repeated elements.

The code below includes a simple test() function that generates all permutations of an ordered numeric string.

#! /usr/bin/env python

''' Find the next permutation in lexicographic order after a given permutation

    This algorithm, due to Narayana Pandita, is from
    https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order

    1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists, 
    the permutation is the last permutation.
    2. Find the largest index k greater than j such that a[j] < a[k].
    3. Swap the value of a[j] with that of a[k].
    4. Reverse the sequence from a[j + 1] up to and including the final element a[n].

    Implemented in Python by PM 2Ring 2015.07.28
'''

import sys

def next_perm(a):
    ''' Advance permutation a to the next one in lexicographic order '''
    n = len(a) - 1
    #1. Find the largest index j such that a[j] < a[j + 1]
    for j in range(n-1, -1, -1):
        if a[j] < a[j + 1]:
            break
    else:
        #This must be the last permutation
        return False

    #2. Find the largest index k greater than j such that a[j] < a[k]
    v = a[j]
    for k in range(n, j, -1):
        if v < a[k]:
            break

    #3. Swap the value of a[j] with that of a[k].
    a[j], a[k] = a[k], a[j]

    #4. Reverse the tail of the sequence
    a[j+1:] = a[j+1:][::-1]

    return True


def test(n):
    ''' Print all permutations of an ordered numeric string (1-based) '''
    a = [str(i) for i in range(1, n+1)]
    i = 0
    while True:
        print('%2d: %s' % (i, ''.join(a)))
        i += 1
        if not next_perm(a):
            break


def main():
    s = sys.argv[1] if len(sys.argv) > 1 else '781623954'
    a = list(s)
    next_perm(a)
    print('%s -> %s' % (s, ''.join(a)))


if __name__ == '__main__':
    #test(4)
    main()

Solution 2:

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.")