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Code Backtracking

Print all binary strings without consecutive 1s

The problem at hand is to generate and print all binary strings of a given length 'k' without consecutive 1s. In other words, we want to find all possible binary strings of length 'k' where no two consecutive characters are '1'. For instance, if k = 3, the valid binary strings would be "000", "001", "010", and "100".

Problem Statement and Description

The problem is to generate and display all binary strings of length 'k' without consecutive 1s. This means that any valid binary string should not contain consecutive '1's. For example, if k = 4, valid strings include "0000", "0010", "0100", and "1000", while "1100" and "0110" are not valid since they contain consecutive '1's.

Example

Let's take a small example to understand the problem better. Suppose k = 3. We want to generate and print all valid binary strings of length 3.

Valid binary strings: "000", "001", "010", "100".

Invalid binary strings: "110", "011", "101", "111".

Idea to Solve

To solve this problem, we can use a recursive approach. We will start building the binary string from left to right, and at each position, we can either place '0' or '1' based on the condition that there should be no consecutive '1's. We will keep track of the generated binary string using an array and pass this array to the recursive function.

Pseudocode

Here's the pseudocode for the solution:

function generateBinaryStrings(record[], start, k):
    if start == k:
        print record
        return
    
    record[start] = '0'
    generateBinaryStrings(record, start + 1, k)
    
    if record[start - 1] != '1':
        record[start] = '1'
        generateBinaryStrings(record, start + 1, k)

function binaryStrings(k):
    if k <= 0:
        return
    
    record = array of size k
    record[0] = '0'
    generateBinaryStrings(record, 1, k)
    
    record[0] = '1'
    generateBinaryStrings(record, 1, k)

Algorithm Explanation

  1. Define a recursive function generateBinaryStrings(record[], start, k) that takes three parameters: the array record to store the generated binary string, the start position where we're currently placing the digit, and k which is the length of the binary string.

  2. Inside generateBinaryStrings, if start reaches k, we have successfully generated a binary string of length k without consecutive '1's, so we print it.

  3. Assign '0' to record[start] and recursively call generateBinaryStrings with start + 1.

  4. Next, check if the previous digit record[start - 1] is not '1'. If it's not, then we can assign '1' to record[start] and recursively call generateBinaryStrings with start + 1.

  5. Define the binaryStrings(k) function which acts as a wrapper. If k is less than or equal to 0, there's no valid string to generate, so we return.

  6. Initialize the record array of size k and start with '0' at the first position. Call generateBinaryStrings(record, 1, k) to generate valid binary strings starting with '0'.

  7. Change the first digit to '1' and again call generateBinaryStrings(record, 1, k) to generate valid binary strings starting with '1'.

Code Solution

/*
    C program for
    Print all binary strings without consecutive 1s
*/
#include <stdio.h>

void solution(char record[], int start, int k)
{
	if (start == k)
	{
		printf(" %s \n", record);
		return;
	}
	if (record[start - 1] == '0')
	{
		record[start] = '0';
		solution(record, start + 1, k);
		// change to 1
		record[start] = '1';
		solution(record, start + 1, k);
	}
	if (record[start - 1] == '1')
	{
		record[start] = '0';
		solution(record, start + 1, k);
	}
}
void binaryString(int k)
{
	// K indicate digit in binary
	if (k <= 0)
	{
		return;
	}
  	// Use to contain result
	char record[k];
  	// Set initial 0
	record[0] = '0';
  	// Display the result which is starting by zeros
	solution(record, 1, k);
	record[0] = '1';
  	// Display the result which is starting by 1s
	solution(record, 1, k);
}
int main(int argc, char const *argv[])
{
	// Test
	binaryString(5);
	return 0;
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
/*
    Java program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	public void solution(String record, int start, int k)
	{
		if (start == k)
		{
			System.out.print(" " + record + " \n");
			return;
		}
		if (record.charAt(start - 1) == '0')
		{
			solution(record + '0', start + 1, k);
			solution(record + '1', start + 1, k);
		}
		if (record.charAt(start - 1) == '1')
		{
			solution(record + '0', start + 1, k);
		}
	}
	public void withoutConsecutive1s(int k)
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		solution("0", 1, k);
		// Display the result which is starting by 1s
		solution("1", 1, k);
	}
	public static void main(String[] args)
	{
		BinaryText task = new BinaryText();
		// Test k = 5
		task.withoutConsecutive1s(5);
	}
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	public: void solution(string record, int start, int k)
	{
		if (start == k)
		{
			cout << " " << record << " \n";
			return;
		}
		if (record[start - 1] == '0')
		{
			this->solution(record  +  '0', start + 1, k);
			this->solution(record  +  '1', start + 1, k);
		}
		if (record[start - 1] == '1')
		{
			this->solution(record  +  '0', start + 1, k);
		}
	}
	void withoutConsecutive1s(int k)
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		this->solution("0", 1, k);
		// Display the result which is starting by 1s
		this->solution("1", 1, k);
	}
};
int main()
{
	BinaryText *task = new BinaryText();
	// Test k = 5
	task->withoutConsecutive1s(5);
	return 0;
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
// Include namespace system
using System;
/*
    Csharp program for
    Print all binary strings without consecutive 1s
*/
public class BinaryText
{
	public void solution(String record, int start, int k)
	{
		if (start == k)
		{
			Console.Write(" " + record + " \n");
			return;
		}
		if (record[start - 1] == '0')
		{
			this.solution(record + '0', start + 1, k);
			this.solution(record + '1', start + 1, k);
		}
		if (record[start - 1] == '1')
		{
			this.solution(record + '0', start + 1, k);
		}
	}
	public void withoutConsecutive1s(int k)
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		this.solution("0", 1, k);
		// Display the result which is starting by 1s
		this.solution("1", 1, k);
	}
	public static void Main(String[] args)
	{
		BinaryText task = new BinaryText();
		// Test k = 5
		task.withoutConsecutive1s(5);
	}
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
package main
import "fmt"
/*
    Go program for
    Print all binary strings without consecutive 1s
*/

func solution(record string, start int, k int) {
	if start == k {
		fmt.Print(" ", record, " \n")
		return
	}
	if record[start - 1] == '0' {
		solution(record + "0", start + 1, k)
		solution(record + "1", start + 1, k)
	}
	if record[start - 1] == '1' {
		solution(record + "0", start + 1, k)
	}
}
func withoutConsecutive1s(k int) {
	// K indicate digit in binary
	if k <= 0 {
		return
	}
	// Display the result which is starting by zeros
	solution("0", 1, k)
	// Display the result which is starting by 1s
	solution("1", 1, k)
}
func main() {
	
	// Test k = 5
	withoutConsecutive1s(5)
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
<?php
/*
    Php program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	public	function solution($record, $start, $k)
	{
		if ($start == $k)
		{
			echo(" ".$record.
				" \n");
			return;
		}
		if ($record[$start - 1] == '0')
		{
			$this->solution($record.strval('0'), $start + 1, $k);
			$this->solution($record.strval('1'), $start + 1, $k);
		}
		if ($record[$start - 1] == '1')
		{
			$this->solution($record.strval('0'), $start + 1, $k);
		}
	}
	public	function withoutConsecutive1s($k)
	{
		// K indicate digit in binary
		if ($k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		$this->solution("0", 1, $k);
		// Display the result which is starting by 1s
		$this->solution("1", 1, $k);
	}
}

function main()
{
	$task = new BinaryText();
	// Test k = 5
	$task->withoutConsecutive1s(5);
}
main();

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
/*
    Node JS program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	solution(record, start, k)
	{
		if (start == k)
		{
			process.stdout.write(" " + record + " \n");
			return;
		}
		if (record.charAt(start - 1) == '0')
		{
			this.solution(record + '0', start + 1, k);
			this.solution(record + '1', start + 1, k);
		}
		if (record.charAt(start - 1) == '1')
		{
			this.solution(record + '0', start + 1, k);
		}
	}
	withoutConsecutive1s(k)
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		this.solution("0", 1, k);
		// Display the result which is starting by 1s
		this.solution("1", 1, k);
	}
}

function main()
{
	var task = new BinaryText();
	// Test k = 5
	task.withoutConsecutive1s(5);
}
main();

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
#    Python 3 program for
#    Print all binary strings without consecutive 1s
class BinaryText :
	def solution(self, record, start, k) :
		if (start == k) :
			print(" ", record ," ")
			return
		
		if (record[start - 1] == '0') :
			self.solution(record + str('0'), start + 1, k)
			self.solution(record + str('1'), start + 1, k)
		
		if (record[start - 1] == '1') :
			self.solution(record + str('0'), start + 1, k)
		
	
	def withoutConsecutive1s(self, k) :
		#  K indicate digit in binary
		if (k <= 0) :
			return
		
		#  Display the result which is starting by zeros
		self.solution("0", 1, k)
		#  Display the result which is starting by 1s
		self.solution("1", 1, k)
	

def main() :
	task = BinaryText()
	#  Test k = 5
	task.withoutConsecutive1s(5)

if __name__ == "__main__": main()

Output

  00000
  00001
  00010
  00100
  00101
  01000
  01001
  01010
  10000
  10001
  10010
  10100
  10101
#    Ruby program for
#    Print all binary strings without consecutive 1s
class BinaryText 
	def solution(record, start, k) 
		if (start == k) 
			print(" ", record ," \n")
			return
		end

		if (record[start - 1] == '0') 
			self.solution(record + '0'.to_s, start + 1, k)
			self.solution(record + '1'.to_s, start + 1, k)
		end

		if (record[start - 1] == '1') 
			self.solution(record + '0'.to_s, start + 1, k)
		end

	end

	def withoutConsecutive1s(k) 
		#  K indicate digit in binary
		if (k <= 0) 
			return
		end

		#  Display the result which is starting by zeros
		self.solution("0", 1, k)
		#  Display the result which is starting by 1s
		self.solution("1", 1, k)
	end

end

def main() 
	task = BinaryText.new()
	#  Test k = 5
	task.withoutConsecutive1s(5)
end

main()

Output

 00000 
 00001 
 00010 
 00100 
 00101 
 01000 
 01001 
 01010 
 10000 
 10001 
 10010 
 10100 
 10101 

import scala.collection.mutable._;
/*
    Scala program for
    Print all binary strings without consecutive 1s
*/
class BinaryText()
{
	def solution(record: String, start: Int, k: Int): Unit = {
		if (start == k)
		{
			print(" " + record + " \n");
			return;
		}
		if (record.charAt(start - 1) == '0')
		{
			solution(record + '0'.toString(), start + 1, k);
			solution(record + '1'.toString(), start + 1, k);
		}
		if (record.charAt(start - 1) == '1')
		{
			solution(record + '0'.toString(), start + 1, k);
		}
	}
	def withoutConsecutive1s(k: Int): Unit = {
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		solution("0", 1, k);
		// Display the result which is starting by 1s
		solution("1", 1, k);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: BinaryText = new BinaryText();
		// Test k = 5
		task.withoutConsecutive1s(5);
	}
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101
import Foundation;
/*
    Swift 4 program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	func solution(_ record: String, _ start: Int, _ k: Int,_ back : String)
	{
		if (start == k)
		{
			print(" ", record ," ");
			return;
		}
		if (back == "0")
		{
			self.solution(record + "0", start + 1, k,"0");
			self.solution(record + "1", start + 1, k,"1");
		}
		if (back == "1")
		{
			self.solution(record + "0", start + 1, k,"0");
		}
	}
	func withoutConsecutive1s(_ k: Int)
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		self.solution("0", 1, k,"0");
		// Display the result which is starting by 1s
		self.solution("1", 1, k,"1");
	}
}
func main()
{
	let task: BinaryText = BinaryText();
	// Test k = 5
	task.withoutConsecutive1s(5);
}
main();

Output

  00000
  00001
  00010
  00100
  00101
  01000
  01001
  01010
  10000
  10001
  10010
  10100
  10101
/*
    Kotlin program for
    Print all binary strings without consecutive 1s
*/
class BinaryText
{
	fun solution(record: String, start: Int, k: Int): Unit
	{
		if (start == k)
		{
			print(" " + record + " \n");
			return;
		}
		if (record.get(start - 1) == '0')
		{
			this.solution(record + "0", start + 1, k);
			this.solution(record + "1", start + 1, k);
		}
		if (record.get(start - 1) == '1')
		{
			this.solution(record + "0", start + 1, k);
		}
	}
	fun withoutConsecutive1s(k: Int): Unit
	{
		// K indicate digit in binary
		if (k <= 0)
		{
			return;
		}
		// Display the result which is starting by zeros
		this.solution("0", 1, k);
		// Display the result which is starting by 1s
		this.solution("1", 1, k);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: BinaryText = BinaryText();
	// Test k = 5
	task.withoutConsecutive1s(5);
}

Output

 00000
 00001
 00010
 00100
 00101
 01000
 01001
 01010
 10000
 10001
 10010
 10100
 10101

Time Complexity

The time complexity of this solution can be analyzed by considering the number of recursive calls. Each recursive call either places '0' or '1' at a specific position, and there are two possible choices at each step. The total number of recursive calls will be 2^k. Therefore, the time complexity is O(2^k), where k is the length of the binary string.

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