Generate all possible permutations of a Number divisible by N

Given a numerical string S, the task is to print all the permutations of the string which are divisible by N.

Examples:

Input: N = 5, S = “125” 
Output: 125 215
Explanation: 
All possible permutations are S are {125, 152, 215, 251, 521, 512}. 
Out of these 6 permutations, only 2 {125, 215} are divisible by N (= 5).

Input: N = 7, S = “4321” 
Output: 4312 4123 3241

Approach: The idea is to generate all possible permutations and for each permutation, check if it is divisible by N or not. For each permutation found to be divisible by N, print them.



Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to Swap two
// characters
void swap_(char& a, char& b)
{
    char temp;
    temp = a;
    a = b;
    b = temp;
}
  
// Function to generate all permutations
// and print the ones that are
// divisible by the N
void permute(char* str, int l, int r, int n)
{
    int i;
  
    if (l == r) {
  
        // Convert string to integer
        int j = atoi(str);
  
        // Check for divisibility
        // and print it
        if (j % n == 0)
            cout << str << endl;
  
        return;
    }
  
    // Print all the permutations
    for (i = l; i < r; i++) {
  
        // Swap characters
        swap_(str[l], str[i]);
  
        // Permute remaining
        // characters
        permute(str, l + 1, r, n);
  
        // Revoke the swaps
        swap_(str[l], str[i]);
    }
}
  
// Driver Code
int main()
{
    char str[100] = "125";
    int n = 5;
    int len = strlen(str);
  
    if (len > 0)
        permute(str, 0, len, n);
  
    return 0;
}

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Java

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// Java program to implement
// the above approach
class GFG{
  
// Function to Swap two
// characters
static void swap_(char []a, int l, int i)
{
    char temp;
    temp = a[l];
    a[l] = a[i];
    a[i] = temp;
}
  
// Function to generate all permutations
// and print the ones that are
// divisible by the N
static void permute(char[] str, int l, 
                         int r, int n)
{
    int i;
  
    if (l == r)
    {
          
        // Convert String to integer
        int j = Integer.valueOf(String.valueOf(str));
  
        // Check for divisibility
        // and print it
        if (j % n == 0)
            System.out.print(String.valueOf(str) + "\n");
  
        return;
    }
  
    // Print all the permutations
    for(i = l; i < r; i++) 
    {
          
        // Swap characters
        swap_(str, l, i);
  
        // Permute remaining
        // characters
        permute(str, l + 1, r, n);
  
        // Revoke the swaps
        swap_(str, l, i);
    }
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "125";
    int n = 5;
    int len = str.length();
  
    if (len > 0)
        permute(str.toCharArray(), 0, len, n);
}
}
  
// This code is contributed by amal kumar choubey 

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Output: 

125
215

Time Complexity: O(N!)
Auxiliary Space: O(N)
 

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Improved By : Amal Kumar Choubey