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# Print all possible strings of length k that can be formed from a set of n characters

• Difficulty Level : Medium
• Last Updated : 28 May, 2022

Given a set of characters and a positive integer k, print all possible strings of length k that can be formed from the given set.

Examples:

```Input:
set[] = {'a', 'b'}, k = 3

Output:
aaa
aab
aba
abb
baa
bab
bba
bbb

Input:
set[] = {'a', 'b', 'c', 'd'}, k = 1
Output:
a
b
c
d```

For a given set of size n, there will be n^k possible strings of length k. The idea is to start from an empty output string (we call it prefix in following code). One by one add all characters to prefix. For every character added, print all possible strings with current prefix by recursively calling for k equals to k-1.

Below is the implementation of above idea :

## C++

 `// C++ program to print all``// possible strings of length k``#include ``using` `namespace` `std;``    `  `// The main recursive method``// to print all possible``// strings of length k``void` `printAllKLengthRec(``char` `set[], string prefix,``                                    ``int` `n, ``int` `k)``{``    ` `    ``// Base case: k is 0,``    ``// print prefix``    ``if` `(k == 0)``    ``{``        ``cout << (prefix) << endl;``        ``return``;``    ``}` `    ``// One by one add all characters``    ``// from set and recursively``    ``// call for k equals to k-1``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``string newPrefix;``        ` `        ``// Next character of input added``        ``newPrefix = prefix + set[i];``        ` `        ``// k is decreased, because``        ``// we have added a new character``        ``printAllKLengthRec(set, newPrefix, n, k - 1);``    ``}` `}` `void` `printAllKLength(``char` `set[], ``int` `k,``int` `n)``{``    ``printAllKLengthRec(set, ``""``, n, k);``}` `// Driver Code``int` `main()``{``    ` `    ``cout << ``"First Test"` `<< endl;``    ``char` `set1[] = {``'a'``, ``'b'``};``    ``int` `k = 3;``    ``printAllKLength(set1, k, 2);``    ` `    ``cout << ``"Second Test\n"``;``    ``char` `set2[] = {``'a'``, ``'b'``, ``'c'``, ``'d'``};``    ``k = 1;``    ``printAllKLength(set2, k, 4);``}` `// This code is contributed``// by Mohit kumar`

## Java

 `// Java program to print all``// possible strings of length k` `class` `GFG {``    ` `// The method that prints all``// possible strings of length k.``// It is mainly a wrapper over``// recursive function printAllKLengthRec()``static` `void` `printAllKLength(``char``[] set, ``int` `k)``{``    ``int` `n = set.length;``    ``printAllKLengthRec(set, ``""``, n, k);``}` `// The main recursive method``// to print all possible``// strings of length k``static` `void` `printAllKLengthRec(``char``[] set,``                               ``String prefix,``                               ``int` `n, ``int` `k)``{``    ` `    ``// Base case: k is 0,``    ``// print prefix``    ``if` `(k == ``0``)``    ``{``        ``System.out.println(prefix);``        ``return``;``    ``}` `    ``// One by one add all characters``    ``// from set and recursively``    ``// call for k equals to k-1``    ``for` `(``int` `i = ``0``; i < n; ++i)``    ``{` `        ``// Next character of input added``        ``String newPrefix = prefix + set[i];``        ` `        ``// k is decreased, because``        ``// we have added a new character``        ``printAllKLengthRec(set, newPrefix,``                                ``n, k - ``1``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``System.out.println(``"First Test"``);``    ``char``[] set1 = {``'a'``, ``'b'``};``    ``int` `k = ``3``;``    ``printAllKLength(set1, k);``    ` `    ``System.out.println(``"\nSecond Test"``);``    ``char``[] set2 = {``'a'``, ``'b'``, ``'c'``, ``'d'``};``    ``k = ``1``;``    ``printAllKLength(set2, k);``}``}`

## Python3

 `# Python 3 program to print all``# possible strings of length k``    ` `# The method that prints all``# possible strings of length k.``# It is mainly a wrapper over``# recursive function printAllKLengthRec()``def` `printAllKLength(``set``, k):` `    ``n ``=` `len``(``set``)``    ``printAllKLengthRec(``set``, "", n, k)` `# The main recursive method``# to print all possible``# strings of length k``def` `printAllKLengthRec(``set``, prefix, n, k):``    ` `    ``# Base case: k is 0,``    ``# print prefix``    ``if` `(k ``=``=` `0``) :``        ``print``(prefix)``        ``return` `    ``# One by one add all characters``    ``# from set and recursively``    ``# call for k equals to k-1``    ``for` `i ``in` `range``(n):` `        ``# Next character of input added``        ``newPrefix ``=` `prefix ``+` `set``[i]``        ` `        ``# k is decreased, because``        ``# we have added a new character``        ``printAllKLengthRec(``set``, newPrefix, n, k ``-` `1``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``print``(``"First Test"``)``    ``set1 ``=` `[``'a'``, ``'b'``]``    ``k ``=` `3``    ``printAllKLength(set1, k)``    ` `    ``print``(``"\nSecond Test"``)``    ``set2 ``=` `[``'a'``, ``'b'``, ``'c'``, ``'d'``]``    ``k ``=` `1``    ``printAllKLength(set2, k)` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# program to print all``// possible strings of length k``using` `System;` `class` `GFG {``    ` `// The method that prints all``// possible strings of length k.``// It is mainly a wrapper over``// recursive function printAllKLengthRec()``static` `void` `printAllKLength(``char``[] ``set``, ``int` `k)``{``    ``int` `n = ``set``.Length;``    ``printAllKLengthRec(``set``, ``""``, n, k);``}` `// The main recursive method``// to print all possible``// strings of length k``static` `void` `printAllKLengthRec(``char``[] ``set``,``                               ``String prefix,``                               ``int` `n, ``int` `k)``{``    ` `    ``// Base case: k is 0,``    ``// print prefix``    ``if` `(k == 0)``    ``{``        ``Console.WriteLine(prefix);``        ``return``;``    ``}` `    ``// One by one add all characters``    ``// from set and recursively``    ``// call for k equals to k-1``    ``for` `(``int` `i = 0; i < n; ++i)``    ``{` `        ``// Next character of input added``        ``String newPrefix = prefix + ``set``[i];``        ` `        ``// k is decreased, because``        ``// we have added a new character``        ``printAllKLengthRec(``set``, newPrefix,``                                ``n, k - 1);``    ``}``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``Console.WriteLine(``"First Test"``);``    ``char``[] set1 = {``'a'``, ``'b'``};``    ``int` `k = 3;``    ``printAllKLength(set1, k);``    ` `    ``Console.WriteLine(``"\nSecond Test"``);``    ``char``[] set2 = {``'a'``, ``'b'``, ``'c'``, ``'d'``};``    ``k = 1;``    ``printAllKLength(set2, k);``}``}` `// This code is contributed by Ajit.`

## Javascript

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Output:

```First Test
aaa
aab
aba
abb
baa
bab
bba
bbb

Second Test
a
b
c
d```

Time complexity: O(nk)

Auxiliary Space: O(k)

The above solution is mainly a generalization of this post.