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Count natural numbers whose all permutation are greater than that number

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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

Explanation:
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.
    stack<int> s;
    for (int i = 1; i <= 9; i++) 
    {
          
        if (i <= n) 
        {
            s.push(i);
            result++;
        }
  
        // take a number from stack and add
        // a digit greater than or equal to 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 Code
int main()
{
    int n = 15;
    cout << countNumber(n) << endl;
    return 0;
}

                    

Java

// Java program to count the number less than N,
// whose all permutation is greater 
// than or equal to the number.
import java.util.Stack;
  
  
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.
        Stack<Integer> s = new Stack<>();
        for (int i = 1; i <= 9; i++) 
        {
  
            if (i <= n)
            {
                s.push(i);
                result++;
            }
  
            // take a number from stack and add
            // a digit greater than or equal to last digit
            // of it.
            while (!s.empty())
            {
                int tp = 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 Code
    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.
    s = []
    for i in range(1, 10):
          
        if (i <= n):
            s.append(i)
            result += 1
  
        # take a number from stack and add
        # a digit greater than or equal to last digit
        # of it.
        while len(s) != 0:
            tp = s[-1]
            s.pop()
            for j in range(tp % 10, 10):
                x = tp * 10 + j
                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

                    

C#

// C# program to count the number less than N,
// whose all permutation is greater than
// or equal to the number.
using System;
using System.Collections.Generic;
  
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.
        Stack<int> s = new Stack<int>();
        for (int i = 1; i <= 9; i++) 
        {
              
            if (i <= n) 
            {
                s.Push(i);
                result++;
            }
  
            // take a number from stack and add
            // a digit greater than or equal to last digit
            // of it.
            while (s.Count != 0) 
            {
                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;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 15;
        Console.WriteLine(countNumber(n));
    }
}
  
// This code is contributed by Rajput-Ji

                    

Javascript

  <script>
  
        // JavaScript program for the above approach
  
        // Return the count of the number having all
        // permutation greater than or equal to the number.
        function countNumber(n)
        {
            let result = 0;
  
            // Pushing 1 to 9 because all number from 1
            // to 9 have this property.
            let s = [];
            for (let i = 1; i <= 9; i++)
            {
  
                if (i <= n)
                {
                    s.push(i);
                    result++;
                }
  
                // take a number from stack and add
                // a digit greater than or equal to last digit
                // of it.
                while (s.length != 0) 
                {
                    let tp = s[s.length - 1];
                    s.pop();
                    for (let j = tp % 10; j <= 9; j++)
                    {
                        let x = tp * 10 + j;
                        if (x <= n)
                        {
                            s.push(x);
                            result++;
                        }
                    }
                }
            }
  
            return result;
        }
  
        // Driven Code
  
        let n = 15;
        document.write(countNumber(n));
  
// This code is contributed by Potta Lokesh
    </script>

                    

Output
14

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

Auxiliary Space: O(x) as it is using stack

 



Last Updated : 11 Sep, 2023
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