Count of substrings of a Binary string containing only 1s

Given a binary string of length N, we need to find out how many substrings of this string contain only 1s.

Examples:

Input: S = “0110111”
Output: 9
Explanation:
There are 9 substring with only 1’s characters. 
“1” comes 5 times. 
“11” comes 3 times. 
“111” comes 1 time.

Input: S = “000”
Output: 0

Approach: The idea is to traverse the binary string and count the consecutive ones in the string. Below is the illustration of the approach:

• Traverse the given binary string from index 0 to length – 1
• Count the number of consecutive “1” till index i.
• For each new character str[i], there will be more substring with all character’s as “1”

Below is the implementation of the above approach:

C++

// C++ implementation to find
// count of substring containing
// only ones

#include <bits/stdc++.h>
using namespace std;

// Function to find the total number
// of substring having only ones
int countOfSubstringWithOnlyOnes(string s)
{
    int res = 0, count = 0;
    for (int i = 0; i < s.length(); i++) {
    count = s[i] == '1' ? count + 1 : 0;
    res = (res + count);
    }
    return res;
}

// Driver Code
int main()
{
    string s = "0110111";
    cout << countOfSubstringWithOnlyOnes(s)
        << endl;
    return 0;
}

Java

// Java implementation to find
// count of substring containing
// only ones
class GFG{

// Function to find the total number
// of substring having only ones
static int countOfSubstringWithOnlyOnes(String s)
{
    int res = 0, count = 0;
    for(int i = 0; i < s.length(); i++)
    {
        count = s.charAt(i) == '1' ? count + 1 : 0;
        res = (res + count);
    }
    return res;
}

// Driver code
public static void main(String[] args)
{
    String s = "0110111";

    System.out.println(countOfSubstringWithOnlyOnes(s));
}
}

// This code is contributed by dewantipandeydp

Python3

# Python3 implementation to find
# count of substring containing
# only ones

# Function to find the total number
# of substring having only ones
def countOfSubstringWithOnlyOnes(s):

    count = 0
    res = 0

    for i in range(0,len(s)):
        if s[i] == '1':
            count = count + 1
        else:
            count = 0;

        res = res + count

    return res

# Driver Code
s = "0110111"
print(countOfSubstringWithOnlyOnes(s))

# This code is contributed by jojo9911

C#

// C# implementation to find count
// of substring containing only ones
using System;

class GFG{

// Function to find the total number
// of substring having only ones
static int countOfSubstringWithOnlyOnes(string s)
{
    int res = 0, count = 0;

    for(int i = 0; i < s.Length; i++)
    {
        count = s[i] == '1' ? count + 1 : 0;
        res = (res + count);
    }
    return res;
}

// Driver code
public static void Main(string[] args)
{
    string s = "0110111";

    Console.Write(countOfSubstringWithOnlyOnes(s));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript implementation to find
// count of substring containing
// only ones

// Function to find the total number
// of substring having only ones
function countOfSubstringWithOnlyOnes(s)
{
    var res = 0, count = 0;
    for (var i = 0; i < s.length; i++) {
    count = s[i] == '1' ? count + 1 : 0;
    res = (res + count);
    }
    return res;
}

// Driver Code
var s = "0110111";
document.write( countOfSubstringWithOnlyOnes(s));

</script>

Output

9

Time Complexity: O(n), where n is the size of the given string
Auxiliary Space: O(1)