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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



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:

C++

// 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 property.
    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;
}

Java

import java.util.Stack;
  
// Java program to count the number less than N,
// whose all permutation is greater than or equal to the number.
class GFG {
// Return the count of the number having all
// permutation greater than or equal to the number.
  
    static int countNumber(int n) {
        int result = 0;
  
        // Pushing 1 to 9 because all number from 1
        // to 9 have this property.
        for (int i = 1; i <= 9; i++) {
            Stack<Integer> s = new Stack<>();
            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.peek();
                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
    public static void main(String[] args) {
        int n = 15;
        System.out.println(countNumber(n));
  
    }
}
  
//this code contributed by Rajput-Ji

Python3

# Python3 program to count the number less 
# than N, whose all permutation is greater 
# than or equal to the number. 
  
# Return the count of the number having 
# all permutation greater than or equal 
# to the number. 
def countNumber(n):
    result = 0
  
    # Pushing 1 to 9 because all number 
    # from 1 to 9 have this property.
    for i in range(1, 10):
        s = [] 
        if (i <= n):
            s.append(i) 
            result += 1
  
        # take a number from stack and add 
        # a digit smaller than last digit 
        # of it. 
        while len(s) != 0:
            tp = s[-1
            s.pop()
            for j in range(tp % 10, 10):
                x = tp * 10 +
                if (x <= n):
                    s.append(x) 
                    result += 1
  
    return result
  
# Driver Code
if __name__ == '__main__':
  
    n = 15
    print(countNumber(n))
  
# This code is contributed by PranchalK


Output:

14

Time Complexity : O(x) where x is number of elements printed in output.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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