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

• Difficulty Level : Medium
• Last Updated : 11 Sep, 2021

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 kvoid 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 Codeint 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 kstatic 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 Codepublic 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 kdef 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 Codeif __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 kusing 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 kstatic 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 Codestatic 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



Output:

First Test
aaa
aab
aba
abb
baa
bab
bba
bbb

Second Test
a
b
c
d

The above solution is mainly a generalization of this post.