There are some natural number whose all permutation is greater than or equal to that number eg. 123, whose all the permutation (123, 231, 321) are greater than or equal to 123.
Given a natural number n, the task is to count all such number from 1 to n.
Input : n = 15. Output : 14 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15 are the numbers whose all permutation is greater than the number itself. So, output 14. Input : n = 100. Output : 54
A simple solution is to run a loop from 1 to n and for every number check if its digits are in non-decreasing order or not.
An efficient solution is based on below observations.
Observation 1: From 1 to 9, all number have this property. So, for n <= 9, output n.
Observation 2: The number whose all permutation is greater than or equal to that number have all their digits in increasing order.
The idea is to push all the number from 1 to 9. Now, pop the top element, say topel and try to make number whose digits are in increasing order and the first digit is topel. To make such numbers, the second digit can be from topel%10 to 9. If this number is less than n, increment the count and push the number in the stack, else ignore.
Below is the implementation of this approach:
Time Complexity : O(x) where x is number of elements printed in output.
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.