# Count natural numbers whose all permutation are greater than that number

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.

Examples:

```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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 C++ implementation of this approach:

```// C++ program to count the number less than N,
// whose all permutation is greater than or equal to the number.
#include<bits/stdc++.h>
using namespace std;

// Return the count of the number having all
// permutation greater than or equal to the number.
int countNumber(int n)
{
int result = 0;

// Pushing 1 to 9 because all number from 1
// to 9 have this propert.
for (int i = 1; i <= 9; i++)
{
stack<int> s;
if (i <= n)
{
s.push(i);
result++;
}

// take a number from stack and add
// a digit smaller than last digit
// of it.
while (!s.empty())
{
int tp = s.top();
s.pop();
for (int j = tp%10; j <= 9; j++)
{
int x = tp*10 + j;
if (x <= n)
{
s.push(x);
result++;
}
}
}
}

return result;
}

// Driven Program
int main()
{
int n = 15;
cout << countNumber(n) << endl;
return 0;
}
```

Output:

```14
```

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.2 Average Difficulty : 3.2/5.0
Based on 16 vote(s)