# 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++

 `// C++ Program to implement ` `// the above approach ` `#include ` `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 = ``"125"``; ` `    ``int` `n = 5; ` `    ``int` `len = ``strlen``(str); ` ` `  `    ``if` `(len > 0) ` `        ``permute(str, 0, len, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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  `

Output:

```125
215
```

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

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Improved By : Amal Kumar Choubey