Print all permutations of a number N greater than itself

Last Updated : 19 Sep, 2022

Given a number N, our task is to print those permutations of integer N which are greater than N.
Examples:

Input: N = 534
Output: 543
Input: N = 324
Output: 342, 423, 432

Approach: To solve this problem, we can obtain all the lexicographically larger permutations of N using next_permutation() method in C++. After getting all such numbers, print them.
For other languages, find the permutations of number N and print the numbers which are greater than N.
Below is the implementation of above approach:

C++

// C++ implementation to print all the
// permutation greater than the integer N

#include <bits/stdc++.h>
using namespace std;

// Function to print all the permutation
// which are greater than N itself
void printPermutation(int N)
{
    int temp = N, count = 0;

    // Iterate and count the
    // number of digits in N
    while (temp > 0) {
        count++;
        temp /= 10;
    }

    // vector to print the
    // permutations of N
    vector<int> num(count);

    // Store digits of N
    // in the vector num
    while (N > 0) {
        num[count-- - 1] = N % 10;
        N = N / 10;
    }

    // Iterate over every permutation of N
    // which is greater than N
    while (next_permutation(
        num.begin(), num.end())) {

        // Print the current permutation of N
        for (int i = 0; i < num.size(); i++)
            cout << num[i];

        cout << "\n";
    }
}

// Driver Code
int main()
{
    int N = 324;

    printPermutation(N);

    return 0;
}

Java

// Java implementation to print all the
// permutation greater than the integer N
import java.util.*;
class GFG{

static void printPermutation(int N) 
{
    int temp = N, count = 0;

    // Iterate and count the
    // number of digits in N
    while (temp > 0) 
    {
      count++;
      temp /= 10;
    }

    // vector to print the
    // permutations of N
    int[] num = new int[count];

    // Store digits of N
    // in the vector num
    while (N > 0)
    {
      num[count-- - 1] = N % 10;
      N = N / 10;
    }

    // Iterate over every permutation of N
    // which is greater than N
    while (next_permutation(num)) 
    {

      // Print the current permutation of N
      for (int i = 0; i < num.length; i++)
        System.out.print(num[i]);

      System.out.print("\n");
    }
  }

  // Function to print all the permutation
  // which are greater than N itself
  static boolean next_permutation(int[] p) 
  {
    for (int a = p.length - 2; a >= 0; --a)
      if (p[a] < p[a + 1])
        for (int b = p.length - 1;; --b)
          if (p[b] > p[a]) 
          {
            int t = p[a];
            p[a] = p[b];
            p[b] = t;
            for (++a, b = p.length - 1; a < b; ++a, --b)
            {
              t = p[a];
              p[a] = p[b];
              p[b] = t;
            }
            return true;
          }
    return false;
  }

// Driver Code
public static void main(String[] args) 
{
  int N = 324;

  printPermutation(N);
}
}

// This code contributed by sapnasingh4991

Python3

# Python3 implementation to print all the
# permutation greater than the integer N

# Function to print all the permutation
# which are greater than N itself
def next_permutation(p):

    for a in range(len(p) - 2, -1, -1):
        if (p[a] < p[a + 1]):
            b = len(p) - 1
            while True:
                if (p[b] > p[a]):
                    t = p[a];
                    p[a] = p[b];
                    p[b] = t;
                    a += 1
                    b = len(p) - 1
                    while a < b:                    
                        t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        a += 1 
                        b -= 1
                    
                    return True;
                b -= 1
        
    return False;



def printPermutation(N):

    temp = N
    count = 0;

    # Iterate and count the
    # number of digits in N
    while (temp > 0):
    
        count += 1
        temp = int(temp / 10)
    

    # vector to print the
    # permutations of N
    num = [0 for _ in range(count)];

    # Store digits of N
    # in the vector num
    while (N > 0):
        count -= 1 
        num[count ] = N % 10;
        N = int(N / 10);
    

    # Iterate over every permutation of N
    # which is greater than N
    while (next_permutation(num)):
    

        # Print the current permutation of N
        print(*num, sep = "")
    

# Driver code
N = 324;

printPermutation(N);

# This code is contributed by phasing17.

// C# implementation to print all the
// permutation greater than the integer N
using System;
class GFG{

static void printPermutation(int N) 
{
    int temp = N, count = 0;

    // Iterate and count the
    // number of digits in N
    while (temp > 0) 
    {
      count++;
      temp /= 10;
    }

    // vector to print the
    // permutations of N
    int[] num = new int[count];

    // Store digits of N
    // in the vector num
    while (N > 0)
    {
      num[count-- - 1] = N % 10;
      N = N / 10;
    }

    // Iterate over every permutation of N
    // which is greater than N
    while (next_permutation(num)) 
    {

      // Print the current permutation of N
      for (int i = 0; i < num.Length; i++)
        Console.Write(num[i]);

      Console.Write("\n");
    }
  }

  // Function to print all the permutation
  // which are greater than N itself
  static bool next_permutation(int[] p) 
  {
    for (int a = p.Length - 2; a >= 0; --a)
      if (p[a] < p[a + 1])
        for (int b = p.Length - 1;; --b)
          if (p[b] > p[a]) 
          {
            int t = p[a];
            p[a] = p[b];
            p[b] = t;
            for (++a, b = p.Length - 1; 
                       a < b; ++a, --b)
            {
              t = p[a];
              p[a] = p[b];
              p[b] = t;
            }
            return true;
          }
    return false;
  }

// Driver Code
public static void Main(String[] args) 
{
  int N = 324;

  printPermutation(N);
}
}

// This code is contributed by Rohit_ranjan

JavaScript

<script>

    // JavaScript implementation to print all the
    // permutation greater than the integer N
    
    function printPermutation(N)
    {
        let temp = N, count = 0;

        // Iterate and count the
        // number of digits in N
        while (temp > 0)
        {
          count++;
          temp = parseInt(temp / 10, 10);
        }

        // vector to print the
        // permutations of N
        let num = new Array(count);
        num.fill(0);

        // Store digits of N
        // in the vector num
        while (N > 0)
        {
          num[count-- - 1] = N % 10;
          N = parseInt(N / 10, 10);
        }

        // Iterate over every permutation of N
        // which is greater than N
        while (next_permutation(num))
        {

          // Print the current permutation of N
          for (let i = 0; i < num.length; i++)
            document.write(num[i]);

          document.write("</br>");
        }
    }

    // Function to print all the permutation
    // which are greater than N itself
    function next_permutation(p)
    {
      for (let a = p.length - 2; a >= 0; --a)
        if (p[a] < p[a + 1])
          for (let b = p.length - 1;; --b)
            if (p[b] > p[a])
            {
              let t = p[a];
              p[a] = p[b];
              p[b] = t;
              for (++a, b = p.length - 1; a < b; ++a, --b)
              {
                t = p[a];
                p[a] = p[b];
                p[b] = t;
              }
              return true;
            }
      return false;
    }

    let N = 324;
 
      printPermutation(N);

</script>

Output:

342
423
432

Print all Unique permutations of a given string.

yatinagg

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Print all permutations of a number N greater than itself

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