Generate all passwords from given character set

Given a set of characters generate all possible passwords from them. This means we should generate all possible permutations of words using the given characters, with repititions and also upto a given length.

Examples:

Input : arr[] = {a, b}, 
          len = 2.
Output :
a b aa ab ba bb

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The solution is to use recursion on the given character array. The idea is to pass all possible lengths and an empty string initially to a helper function. In the helper function we keep appending all the characters one by one to the current string and recur to fill the remaining string till the desired length is reached.

It can be better visualized using the below recursion tree:


        (a, b)
         /   \
        a     b
       / \   / \
      aa  ab ba bb

Following is the implementation of the above method.

C++

// C++ program to generate all passwords for given characters 
#include <bits/stdc++.h> 
using namespace std; 
  
// int cnt; 
  
// Recursive helper function, adds/removes characters 
// until len is reached 
void generate(char* arr, int i, string s, int len) 
{ 
    // base case 
    if (i == 0) // when len has been reached 
    { 
        // print it out 
        cout << s << "\n"; 
        // cnt++; 
        return; 
    } 
  
    // iterate through the array 
    for (int j = 0; j < len; j++) { 
  
        // Create new string with next character 
        // Call generate again until string has 
        // reached its len 
        string appended = s + arr[j]; 
        generate(arr, i - 1, appended, len); 
    } 
  
    return; 
} 
  
// function to generate all possible passwords 
void crack(char* arr, int len) 
{ 
    // call for all required lengths 
    for (int i = 1; i <= len; i++) { 
        generate(arr, i, "", len); 
    } 
} 
  
// driver function 
int main() 
{ 
    char arr[] = { 'a', 'b', 'c' }; 
    int len = sizeof(arr) / sizeof(arr[0]); 
    crack(arr, len); 
  
    //cout << cnt << endl; 
    return 0; 
} 
// This code is contributed by Satish Srinivas. 

Java

// Java program to generate all passwords for given characters 
import java.util.*; 
  
class GFG 
{ 
  
    // int cnt; 
    // Recursive helper function, adds/removes characters 
    // until len is reached 
    static void generate(char[] arr, int i, String s, int len) 
    { 
        // base case 
        if (i == 0) // when len has been reached 
        { 
            // print it out 
            System.out.println(s); 
              
            // cnt++; 
            return; 
        } 
  
        // iterate through the array 
        for (int j = 0; j < len; j++) 
        { 
  
            // Create new string with next character 
            // Call generate again until string has 
            // reached its len 
            String appended = s + arr[j]; 
            generate(arr, i - 1, appended, len); 
        } 
  
        return; 
    } 
  
    // function to generate all possible passwords 
    static void crack(char[] arr, int len) 
    { 
        // call for all required lengths 
        for (int i = 1; i <= len; i++) 
        { 
            generate(arr, i, "", len); 
        } 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        char arr[] = {'a', 'b', 'c'}; 
        int len = arr.length; 
        crack(arr, len); 
    } 
  
} 
  
// This code has been contributed by 29AjayKumar 

Python 3

# Python3 program to  
# generate all passwords 
# for given characters 
  
# Recursive helper function,  
# adds/removes characters 
# until len is reached 
def generate(arr, i, s, len): 
  
    # base case 
    if (i == 0): # when len has 
                 # been reached 
      
        # print it out 
        print(s) 
        return
      
    # iterate through the array 
    for j in range(0, len): 
  
        # Create new string with  
        # next character Call  
        # generate again until  
        # string has reached its len 
        appended = s + arr[j] 
        generate(arr, i - 1, appended, len) 
  
    return
  
# function to generate  
# all possible passwords 
def crack(arr, len): 
  
    # call for all required lengths 
    for i in range(1 , len + 1):  
        generate(arr, i, "", len) 
      
# Driver Code 
arr = ['a', 'b', 'c' ] 
len = len(arr) 
crack(arr, len) 
  
# This code is contributed by Smita. 

C#

// C# program to generate all passwords for given characters 
using System; 
      
class GFG 
{ 
  
    // int cnt; 
    // Recursive helper function, adds/removes characters 
    // until len is reached 
    static void generate(char[] arr, int i, String s, int len) 
    { 
        // base case 
        if (i == 0) // when len has been reached 
        { 
            // print it out 
            Console.WriteLine(s); 
              
            // cnt++; 
            return; 
        } 
  
        // iterate through the array 
        for (int j = 0; j < len; j++) 
        { 
  
            // Create new string with next character 
            // Call generate again until string has 
            // reached its len 
            String appended = s + arr[j]; 
            generate(arr, i - 1, appended, len); 
        } 
  
        return; 
    } 
  
    // function to generate all possible passwords 
    static void crack(char[] arr, int len) 
    { 
        // call for all required lengths 
        for (int i = 1; i <= len; i++) 
        { 
            generate(arr, i, "", len); 
        } 
    } 
  
    // Driver code 
    public static void Main(String[] args) 
    { 
        char []arr = {'a', 'b', 'c'}; 
        int len = arr.Length; 
        crack(arr, len); 
    } 
  
} 
  
/* This code contributed by PrinciRaj1992 */

Output:

a
b
c
aa
ab
ac
ba
bb
bc
ca
cb
cc
aaa
aab
aac
aba
abb
abc
aca
acb
acc
baa
bab
bac
bba
bbb
bbc
bca
bcb
bcc
caa
cab
cac
cba
cbb
cbc
cca
ccb
ccc

If we want to see the count of the words, we can uncomment the lines having cnt variable in the code. We can observe that it comes out to be $n^1 + n^2 +..+ n^n$ , where n = len. Thus the time complexity of the program is also $O(n * n^n)$ , hence exponential. We can also check for a particular password
while generating and break the loop and return if it is found. We can also include other symbols to be generated and if needed remove duplicates by preprocessing the input using a HashTable.

This article is contributed by Satish Srinivas. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

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