Count of groups of consecutive 1s in a given Binary String

0 0 4 minutes read

Given a binary string S of size N, the task is to find the number of groups of 1s only in the string S.

Examples:

Input: S = “100110111”, N = 9
Output: 3
Explanation:
The following groups are of 1s only:

Group over the range [0, 0] which is equal to “1”.

Group over the range [3, 4] which is equal to “11”.

Group over the range [6, 8] which is equal to “111”.

Therefore, there are a total of 3 groups of 1s only.

Input: S = “0101”
Output: 2

Approach: The problem can be solved by iterating over the characters of the string. Follow the steps below to solve the problem:

Initialize a variable, say count as 0, which stores the number of substrings of 1s in S.
Initialize a stack say st to store the substring before an index of 1s only.
Iterate over the characters of the string S, using the variable i and do the following:
- If the current character is 1, push 1 into stack st.
- Otherwise, If st is not empty, increment count by 1. Else Clear st.
If st is not empty, increment count by 1, i.e If there is a suffix of 1s.
Finally, print the total count obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number of the
// groups of 1s only in the binary
// string
int groupsOfOnes(string S, int N)
{
    // Stores number of groups of 1s
    int count = 0;
 
    // Initialization of the stack
    stack<int> st;
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
        // If S[i] is '1'
        if (S[i] == '1')
            st.push(1);
        // Otherwise
        else {
            // If st is empty
            if (!st.empty()) {
                count++;
                while (!st.empty()) {
                    st.pop();
                }
            }
        }
    }
    // If st is not empty
    if (!st.empty())
        count++;
 
    // Return answer
    return count;
}
// Driver code
int main()
{
    // Input
    string S = "100110111";
    int N = S.length();
 
    // Function call
    cout << groupsOfOnes(S, N) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.Stack;
 
class GFG{
 
// Function to find the number of the
// groups of 1s only in the binary
// string
static int groupsOfOnes(String S, int N)
{
     
    // Stores number of groups of 1s
    int count = 0;
 
    // Initialization of the stack
    Stack<Integer> st = new Stack<>();
 
    // Traverse the string S
    for(int i = 0; i < N; i++) 
    {
         
        // If S[i] is '1'
        if (S.charAt(i) == '1')
            st.push(1);
             
        // Otherwise
        else
        {
             
            // If st is empty
            if (!st.empty())
            {
                count++;
                while (!st.empty()) 
                {
                    st.pop();
                }
            }
        }
    }
     
    // If st is not empty
    if (!st.empty())
        count++;
 
    // Return answer
    return count;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    String S = "100110111";
    int N = S.length();
 
    // Function call
    System.out.println(groupsOfOnes(S, N));
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to find the number of the
# groups of 1s only in the binary
# string
def groupsOfOnes(S, N):
     
    # Stores number of groups of 1s
    count = 0
 
    # Initialization of the stack
    st = []
 
    # Traverse the string S
    for i in range(N):
         
        # If S[i] is '1'
        if (S[i] == '1'):
            st.append(1)
             
        # Otherwise
        else:
             
            # If st is empty
            if (len(st) > 0):
                count += 1
                while (len(st) > 0):
                    del st[-1]
                     
    # If st is not empty
    if (len(st)):
        count += 1
 
    # Return answer
    return count
     
# Driver code
if __name__ == '__main__':
     
    # Input
    S = "100110111"
    N = len(S)
     
    # Function call
    print(groupsOfOnes(S, N))
     
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the number of the
// groups of 1s only in the binary
// string
static int groupsOfOnes(string S, int N)
{
   
    // Stores number of groups of 1s
    int count = 0;
 
    // Initialization of the stack
    Stack<int> st = new Stack<int>();
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
        // If S[i] is '1'
        if (S[i] == '1')
            st.Push(1);
        // Otherwise
        else {
            // If st is empty
            if (st.Count > 0) {
                count++;
                while (st.Count > 0) {
                    st.Pop();
                }
            }
        }
    }
   
    // If st is not empty
    if (st.Count > 0)
        count++;
 
    // Return answer
    return count;
}
   
// Driver code
public static void Main()
{
    // Input
    string S = "100110111";
    int N = S.Length;
 
    // Function call
    Console.Write(groupsOfOnes(S, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>
 
        // JavaScript program for the above approach
 
        // Function to find the number of the
        // groups of 1s only in the binary
        // string
        function groupsOfOnes(S, N) {
            // Stores number of groups of 1s
            let count = 0;
 
            // Initialization of the stack
            var st = [];
 
            // Traverse the string S
            for (let i = 0; i < N; i++) {
 
                // If S[i] is '1'
                if (S[i] == '1')
                    st.push(1);
                // Otherwise
                else {
                    // If st is empty
                    if (st.length != 0) {
                        count++;
                        while (st.length != 0) {
                            st.pop();
                        }
                    }
                }
            }
            // If st is not empty
            if (st.length != 0)
                count++;
 
            // Return answer
            return count;
        }
        // Driver code
 
        // Input
        var S = "100110111";
        let N = S.length;
 
        // Function call
        document.write(groupsOfOnes(S, N));
 
// This code is contributed by Potta Lokesh
 
</script>

Output

Time Complexity: O(N)
Auxiliary Space: O(N)

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!