Generate all binary strings of length n with sub-string “01” appearing exactly twice

Given an integer N, the task is to generate all possible binary strings of length N which contain “01” as the sub-string exactly twice.

Examples:

Input: N = 4
Output:
0101
“0101” is the only binary string of length 4
that contains “01” exactly twice as the sub-string.

Input: N = 5
Output:
00101
01001
01010
01011
01101
10101

Approach: This problem can solved using backtracking. To generate a binary string, we implement a function that generate each bit at a time, update the state of the binary string (current length, number of occurrences of the pattern). Then call the function recursively, and according to the current state of the binary string, the function will decide how to generate the next bit or print out the binary string (if the problem’s requirement is met).

For this problem, backtracking strategy looks like we generate a binary tree with each node can have either value 0 or 1.
For example, with N = 4, the tree will look like:

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
#include <stdlib.h>
using namespace std;
  
// Utility function to print the given binary string
void printBinStr(int* str, int len)
{
    for (int i = 0; i < len; i++) {
        cout << str[i];
    }
    cout << endl;
}
  
// This function will be called recursively
// to generate the next bit for given
// binary string according to its current state
void generateBinStr(int* str, int len, int currlen,
                    int occur, int nextbit)
{
  
    // Base-case: if the generated binary string
    // meets the required length and the pattern "01"
    // appears twice
    if (currlen == len) {
  
        // nextbit needs to be  0 because each time
        // we call the function recursively,
        // we call 2 times for 2 cases:
        // next bit is 0 or 1
        // The is to assure that the binary
        // string is printed one time only
        if (occur == 2 && nextbit == 0)
            printBinStr(str, len);
        return;
    }
  
    // Generate the next bit for str
    // and call recursive
    if (currlen == 0) {
  
        // Assign first bit
        str[0] = nextbit;
  
        // The next generated bit will wither be 0 or 1
        generateBinStr(str, len, currlen + 1, occur, 0);
        generateBinStr(str, len, currlen + 1, occur, 1);
    }
    else {
  
        // If pattern "01" occurrence is < 2
        if (occur < 2) {
  
            // Set next bit
            str[currlen] = nextbit;
  
            // If pattern "01" appears then
            // increase the occurrence of pattern
            if (str[currlen - 1] == 0 && nextbit == 1) {
                occur += 1;
            }
            generateBinStr(str, len, currlen + 1, occur, 0);
            generateBinStr(str, len, currlen + 1, occur, 1);
  
            // Else pattern "01" occurrence equals 2
        }
        else {
  
            // If previous bit is 0 then next bit cannot be 1
            if (str[currlen - 1] == 0 && nextbit == 1) {
                return;
  
                // Otherwise
            }
            else {
                str[currlen] = nextbit;
                generateBinStr(str, len, currlen + 1, occur, 0);
                generateBinStr(str, len, currlen + 1, occur, 1);
            }
        }
    }
}
  
// Driver code
int main()
{
  
    int n = 5;
  
    // Length of the resulting strings
    // must be at least 4
    if (n < 4)
        cout << -1;
    else {
        int* str = new int[n];
  
        // Generate all binary strings of length n
        // with sub-string "01" appearing twice
        generateBinStr(str, n, 0, 0, 0);
        generateBinStr(str, n, 0, 0, 1);
    }
  
    return 0;
}

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Java

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// Java implementation of the above approach 
class GFG 
{
  
    // Utility function to print the given binary string
    static void printBinStr(int[] str, int len) 
    {
        for (int i = 0; i < len; i++)
        {
            System.out.print(str[i]);
        }
        System.out.println();
    }
  
    // This function will be called recursively
    // to generate the next bit for given
    // binary string according to its current state
    static void generateBinStr(int[] str, int len, int currlen,
                                    int occur, int nextbit) 
    {
  
        // Base-case: if the generated binary string
        // meets the required length and the pattern "01"
        // appears twice
        if (currlen == len) 
        {
  
            // nextbit needs to be 0 because each time
            // we call the function recursively,
            // we call 2 times for 2 cases:
            // next bit is 0 or 1
            // The is to assure that the binary
            // string is printed one time only
            if (occur == 2 && nextbit == 0)
            {
                printBinStr(str, len);
            }
            return;
        }
  
        // Generate the next bit for str
        // and call recursive
        if (currlen == 0
        {
  
            // Assign first bit
            str[0] = nextbit;
  
            // The next generated bit will wither be 0 or 1
            generateBinStr(str, len, currlen + 1, occur, 0);
            generateBinStr(str, len, currlen + 1, occur, 1);
        } else // If pattern "01" occurrence is < 2
        if (occur < 2
        {
  
            // Set next bit
            str[currlen] = nextbit;
  
            // If pattern "01" appears then
            // increase the occurrence of pattern
            if (str[currlen - 1] == 0 && nextbit == 1
            {
                occur += 1;
            }
            generateBinStr(str, len, currlen + 1, occur, 0);
            generateBinStr(str, len, currlen + 1, occur, 1);
  
            // Else pattern "01" occurrence equals 2
        } else // If previous bit is 0 then next bit cannot be 1
        if (str[currlen - 1] == 0 && nextbit == 1
        {
            return;
  
            // Otherwise
        
        else 
        {
            str[currlen] = nextbit;
            generateBinStr(str, len, currlen + 1, occur, 0);
            generateBinStr(str, len, currlen + 1, occur, 1);
        }
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int n = 5;
  
        // Length of the resulting strings
        // must be at least 4
        if (n < 4
        {
            System.out.print(-1);
        
        else
        {
            int[] str = new int[n];
  
            // Generate all binary strings of length n
            // with sub-string "01" appearing twice
            generateBinStr(str, n, 0, 0, 0);
            generateBinStr(str, n, 0, 0, 1);
        }
    }
}
  
// This code has been contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
  
# Utility function to print the 
# given binary string
def printBinStr(string, length):
  
    for i in range(0, length): 
        print(string[i], end = "")
      
    print()
  
# This function will be called recursively
# to generate the next bit for given
# binary string according to its current state
def generateBinStr(string, length, currlen,
                            occur, nextbit):
  
    # Base-case: if the generated binary
    # string meets the required length and
    # the pattern "01" appears twice
    if currlen == length:
  
        # nextbit needs to be 0 because each
        # time we call the function recursively,
        # we call 2 times for 2 cases:
        # next bit is 0 or 1
        # The is to assure that the binary
        # string is printed one time only
        if occur == 2 and nextbit == 0:
            printBinStr(string, length)
        return
  
    # Generate the next bit for
    # str and call recursive
    if currlen == 0
  
        # Assign first bit
        string[0] = nextbit
  
        # The next generated bit will 
        # either be 0 or 1
        generateBinStr(string, length,
                       currlen + 1, occur, 0)
        generateBinStr(string, length, 
                       currlen + 1, occur, 1)
      
    else:
  
        # If pattern "01" occurrence is < 2
        if occur < 2
  
            # Set next bit
            string[currlen] = nextbit
  
            # If pattern "01" appears then
            # increase the occurrence of pattern
            if string[currlen - 1] == 0 and nextbit == 1:
                occur += 1
              
            generateBinStr(string, length, 
                           currlen + 1, occur, 0)
            generateBinStr(string, length, 
                           currlen + 1, occur, 1)
  
            # Else pattern "01" occurrence equals 2
          
        else:
  
            # If previous bit is 0 then next bit cannot be 1
            if string[currlen - 1] == 0 and nextbit == 1
                return
  
                # Otherwise
              
            else:
                string[currlen] = nextbit
                generateBinStr(string, length, 
                               currlen + 1, occur, 0)
                generateBinStr(string, length,
                               currlen + 1, occur, 1)
  
# Driver code
if __name__ == "__main__":
  
    n = 5
  
    # Length of the resulting strings
    # must be at least 4
    if n < 4:
        print(-1)
    else:
        string = [None] * n
  
        # Generate all binary strings of length n
        # with sub-string "01" appearing twice
        generateBinStr(string, n, 0, 0, 0)
        generateBinStr(string, n, 0, 0, 1)
      
# This code is contributed by Rituraj Jain

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C#

// C# implementation of the above approach
using System;

class GFG
{

// Utility function to print the given binary string
static void printBinStr(int[] str, int len)
{
for (int i = 0; i < len; i++) { Console.Write(str[i]); } Console.Write("\n"); } // This function will be called recursively // to generate the next bit for given // binary string according to its current state static void generateBinStr(int[] str, int len, int currlen, int occur, int nextbit) { // Base-case: if the generated binary string // meets the required length and the pattern "01" // appears twice if (currlen == len) { // nextbit needs to be 0 because each time // we call the function recursively, // we call 2 times for 2 cases: // next bit is 0 or 1 // The is to assure that the binary // string is printed one time only if (occur == 2 && nextbit == 0) { printBinStr(str, len); } return; } // Generate the next bit for str // and call recursive if (currlen == 0) { // Assign first bit str[0] = nextbit; // The next generated bit will wither be 0 or 1 generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } else // If pattern "01" occurrence is < 2 if (occur < 2) { // Set next bit str[currlen] = nextbit; // If pattern "01" appears then // increase the occurrence of pattern if (str[currlen - 1] == 0 && nextbit == 1) { occur += 1; } generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); // Else pattern "01" occurrence equals 2 } else // If previous bit is 0 then next bit cannot be 1 if (str[currlen - 1] == 0 && nextbit == 1) { return; // Otherwise } else { str[currlen] = nextbit; generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } } // Driver code public static void Main(String[] args) { int n = 5; // Length of the resulting strings // must be at least 4 if (n < 4) { Console.Write(-1); } else { int[] str = new int[n]; // Generate all binary strings of length n // with sub-string "01" appearing twice generateBinStr(str, n, 0, 0, 0); generateBinStr(str, n, 0, 0, 1); } } } // This code is contributed by Princi Singh [tabbyending]

Output:

00101
01001
01010
01011
01101
10101


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