Generate all binary strings without consecutive 1’s

• Difficulty Level : Medium
• Last Updated : 18 Jan, 2022

Given an integer K. Task is Print All binary string of size K (Given number).
Examples:

Input : K = 3
Output : 000 , 001 , 010 , 100 , 101

Input : K  = 4
Output :0000 0001 0010 0100 0101 1000 1001 1010

Idea behind that is IF string ends with ‘1’ then we put only ‘0’ at the end. IF string ends with ‘0’ then we put both ‘0’ and ‘1’ at the end of string for generating new string.
Below is algorithm

K : size of string
First We Generate All string starts with '0'
initialize n = 1 .
GenerateALLString ( K  , Str , n )
a. IF n == K
PRINT str.
b. IF previous character is '1' :: str[n-1] == '1'
put str[n] = '0'
GenerateAllString ( K , str , n+1 )
c. IF previous character is '0' :: str[n-1] == '0'
First We Put zero at end and call function
PUT  str[n] = '0'
GenerateAllString ( K , str , n+1 )
PUT  str[n] = '1'
GenerateAllString ( K , str , n+1 )

Second Generate all binary string starts with '1'
DO THE SAME PROCESS

Below is the recursive implementation:

C++

 // C++ program to Generate// all binary string without// consecutive 1's of size K#includeusing namespace std ; // A utility function generate all string without// consecutive 1'sof size Kvoid generateAllStringsUtil(int K, char str[], int n){         // Print binary string without consecutive 1's    if (n  == K)    {                 // Terminate binary string        str[n] = '\0' ;        cout << str << " ";        return ;    }     // If previous character is '1' then we put    // only 0 at end of string    //example str = "01" then new string be "010"    if (str[n-1] == '1')    {        str[n] = '0';        generateAllStringsUtil (K , str , n+1);    }     // If previous character is '0' than we put    // both '1' and '0' at end of string    // example str = "00" then    // new string "001" and "000"    if (str[n-1] == '0')    {        str[n] = '0';        generateAllStringsUtil(K, str, n+1);        str[n] = '1';        generateAllStringsUtil(K, str, n+1) ;    }} // Function generate all binary string without// consecutive 1'svoid generateAllStrings(int K ){    // Base case    if (K <= 0)        return ;     // One by one stores every    // binary string of length K    char str[K];     // Generate all Binary string    // starts with '0'    str = '0' ;    generateAllStringsUtil ( K , str , 1 ) ;     // Generate all Binary string    // starts with '1'    str = '1' ;    generateAllStringsUtil ( K , str , 1 );} // Driver program to test above functionint main(){    int K = 3;    generateAllStrings (K) ;    return 0;}

Java

 // Java program to Generate all binary string without// consecutive 1's of size Kimport java.util.*;import java.lang.*; public class BinaryS {     // Array conversion to String--    public static String toString(char[] a) {        String string = new String(a);        return string;    }     static void generate(int k, char[] ch, int n) {               // Base Condition when we        // reached at the end of Array**        if (n == k) {                       // Printing the Generated String**            // Return to the previous case*            System.out.print(toString(ch)+" ");            return;         }               // If the first Character is        //Zero then adding**        if (ch[n - 1] == '0') {            ch[n] = '0';            generate(k, ch, n + 1);            ch[n] = '1';            generate(k, ch, n + 1);        }               // If the Character is One        // then add Zero to next**        if (ch[n - 1] == '1') {             ch[n] = '0';                       // Calling Recursively for the            // next value of Array            generate(k, ch, n + 1);         }    }     static void fun(int k) {         if (k <= 0) {            return;        }         char[] ch = new char[k];               // Initializing first character to Zero        ch = '0';               // Generating Strings starting with Zero--        generate(k, ch, 1);         // Initialized first Character to one--        ch = '1';        generate(k, ch, 1);     }     public static void main(String args[]) {         int k = 3;               //Calling function fun with argument k        fun(k);               //This code is Contributed by Praveen Tiwari    }}

Python3

 # Python3 program to Generate all binary string# without consecutive 1's of size K # A utility function generate all string without# consecutive 1'sof size Kdef generateAllStringsUtil(K, str, n):         # print binary string without consecutive 1's    if (n == K):                 # terminate binary string        print(*str[:n], sep = "", end = " ")        return         # if previous character is '1' then we put    # only 0 at end of string    # example str = "01" then new string be "000"    if (str[n-1] == '1'):        str[n] = '0'        generateAllStringsUtil (K, str, n + 1)             # if previous character is '0' than we put    # both '1' and '0' at end of string    # example str = "00" then new string "001" and "000"    if (str[n-1] == '0'):        str[n] = '0'        generateAllStringsUtil(K, str, n + 1)        str[n] = '1'        generateAllStringsUtil(K, str, n + 1)         # function generate all binary string without# consecutive 1'sdef generateAllStrings(K):         # Base case    if (K <= 0):        return         # One by one stores every    # binary string of length K    str =  * K         # Generate all Binary string starts with '0'    str = '0'    generateAllStringsUtil (K, str, 1)         # Generate all Binary string starts with '1'    str = '1'    generateAllStringsUtil (K, str, 1) # Driver codeK = 3generateAllStrings (K) # This code is contributed by SHUBHAMSINGH10

C#

 // C# program to Generate// all binary string without// consecutive 1's of size Kusing System;class GFG {   // Array conversion to String--  static string toString(char[] a) {    string String = new string(a);    return String;  }   static void generate(int k, char[] ch, int n) {     // Base Condition when we    // reached at the end of Array**    if (n == k) {       // Printing the Generated String**      // Return to the previous case*      Console.Write(toString(ch)+" ");      return;     }     // If the first Character is    //Zero then adding**    if (ch[n - 1] == '0') {      ch[n] = '0';      generate(k, ch, n + 1);      ch[n] = '1';      generate(k, ch, n + 1);    }     // If the Character is One    // then add Zero to next**    if (ch[n - 1] == '1') {       ch[n] = '0';       // Calling Recursively for the      // next value of Array      generate(k, ch, n + 1);     }  }   static void fun(int k)  {    if (k <= 0)    {      return;    }    char[] ch = new char[k];     // Initializing first character to Zero    ch = '0';     // Generating Strings starting with Zero--    generate(k, ch, 1);     // Initialized first Character to one--    ch = '1';    generate(k, ch, 1);   }   // Driver code  static void Main()  {    int k = 3;     //Calling function fun with argument k    fun(k);  }} // This code is contributed by divyeshrabadiya07.

Javascript



Output:

000 001 010 100 101

This problem is solved by using a recursion tree having two possibilities 0 or 1 just like selecting elements in a subsequence.

So we can also implement above approach using boolean array as well.

C++14

 #include using namespace std; void All_Binary_Strings(bool arr[],int num,int r){    if(r==num)    {        for(int i=0;i

The above approach can also be solved using string. It does not have any effect on complexity but string handling, printing and operating is easy.

C++14

 #include using namespace std; void All_Binary_Strings(string str,int num){    int len=str.length();    if(len==num)    {        cout<

Time Complexity: O(2^n)

Auxiliary Space: O(n)

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