Skip to content
Related Articles

Related Articles

Improve Article

Count binary strings of length same as given string after removal of substrings “01” and “00” that consists of at least one ‘1’

  • Last Updated : 05 May, 2021

Given a binary string S, the task is to count total binary strings consisting of at least one ‘1’ of length equal to the length of the given string after repeatedly removing all occurrences of substrings “10” and “00” from the given string.

Examples:

Input: S = “111”
Output: 7
Explanation: Since there are no occurrences of “10” or “01” in the given string, length of the remaining string is 3. All possible binary strings of length 3 consisting of at least one set bit are {“001”, “010”, “011”, “100”, “101”, “110”, “111”}

Input: S = “0101”
Output: 3
Explanation: After deleting “10”, S = “01”. Therefore, length of remaining string is 2. Strings of length 2 consisting of at least one set bit are {“01”, “10”, “11”}

Approach: The idea is to calculate the length of the given string after removing all substrings of the form “10” and “00” from it. Considering the remaining; length of the string to be N, the total number of strings consisting of at least one set bit will be equal to 2N-1.



Follow the below steps to solve the problem:

  1. Initialize a variable, say count.
  2. Iterate over the characters of the string S. For each character, check if it is ‘0’ and the count is greater than 0 or not. If found to be true, decrement the count by 1.
  3. Otherwise, if the current character is ‘1’, increment count by 1.
  4. After complete traversal of the string, print 2count – 1 as the required answer.

 Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the strings
// consisting of at least 1 set bit
void countString(string S)
{
    // Initialize count
    long long count = 0;
 
    // Iterate through string
    for (auto it : S) {
 
        if (it == '0' and count > 0) {
            count--;
        }
        else {
            count++;
        }
    }
 
    // The answer is 2^N-1
    cout << ((1 << count) - 1) << "\n";
}
 
// Driver Code
int main()
{
 
    // Given string
    string S = "1001";
 
    // Function call
    countString(S);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
   
class GFG{
   
// Function to count the strings
// consisting of at least 1 set bit
static void countString(String S)
{
     
    // Initialize count
    int count = 0;
     
    // Iterate through string
    for(char it : S.toCharArray())
    {
        if (it == '0' && count > 0)
        {
            count--;
        }
        else
        {
            count++;
        }
    }
  
    // The answer is 2^N-1
    System.out.print((1 << count) - 1);
}
   
// Driver Code
public static void main(String[] args)
{
     
    // Given string
    String S = "1001";
  
    // Function call
    countString(S);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python3 program for the
# above approach 
 
# Function to count the
# strings consisting of
# at least 1 set bit
def countString(S):
   
    # Initialize count
    count = 0
     
    # Iterate through
    # string
    for i in S:
        if (i == '0' and
            count > 0):           
            count -= 1
         
        else:           
            count += 1
 
    # The answer is 2^N-1    
    print((1 << count) - 1
 
# Driver Code
if __name__ == "__main__":
 
    # Given string   
    S = "1001"
     
    # Function call   
    countString(S)
 
# This code is contributed by Virusbuddah_

C#




// C# program for the above approach
using System;
   
class GFG{
   
// Function to count the strings
// consisting of at least 1 set bit
static void countString(string S)
{
     
    // Initialize count
    int count = 0;
     
    // Iterate through string
    foreach(char it in S)
    {
        if (it == '0' && count > 0)
        {
            count--;
        }
        else
        {
            count++;
        }
    }
  
    // The answer is 2^N-1
    Console.Write((1 << count) - 1);
}
   
// Driver Code
public static void Main(string[] args)
{
     
    // Given string
    string S = "1001";
  
    // Function call
    countString(S);
}
}
 
// This code is contributed by chitranayal

Javascript




<script>
// javascript program for the above approach
 
    // Function to count the strings
    // consisting of at least 1 set bit
    function countString( S) {
 
        // Initialize count
        var count = 0;
 
        // Iterate through string
        for (var it =0;it< S.length; it++) {
            if (S[it] == '0' && count > 0) {
                count--;
            } else {
                count++;
            }
        }
 
        // The answer is 2^N-1
        document.write((1 << count) - 1);
    }
 
    // Driver Code
     
 
        // Given string
        var S = "1001";
 
        // Function call
        countString(S);
 
// This code contributed by umadevi9616
</script>
Output: 
3

 

Time Complexity: O(L) where L is the length of the string. 
Auxiliary Space: O(1) 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :