Number of sub-strings in a given binary string divisible by 2

Given binary string str of length N, the task is to find the count of substrings of str which are divisible by 2. Leading zeros in a substring are allowed.

Examples:

Input: str = “101” 
Output: 2 
“0” and “10” are the only substrings 
which are divisible by 2.

Input: str = “10010” 
Output: 10 

Naive approach: A naive approach will be to generate all possible substrings and check if they are divisible by 2. The time complexity for this will be O(N³).

Efficient approach: It can be observed that any binary number is divisible by 2 only if it ends with a 0. Now, the task is to just count the number of substrings ending with 0. So, for every index i such that str[i] = ‘0’, find the number of substrings ending at i. This value is equal to (i + 1) (0-based indexing). Thus, the final answer will be equal to the summation of (i + 1) for all i such that str[i] = ‘0’.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of the required substrings
int countSubStr(string str, int len)
{
    // To store the final answer
    int ans = 0;

    // Loop to find the answer
    for (int i = 0; i < len; i++) {

        // Condition to update the answer
        if (str[i] == '0')
            ans += (i + 1);
    }

    return ans;
}

// Driver code
int main()
{
    string str = "10010";
    int len = str.length();

    cout << countSubStr(str, len);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function to return the count
    // of the required substrings
    static int countSubStr(String str, int len)
    {
        // To store the final answer
        int ans = 0;

        // Loop to find the answer
        for (int i = 0; i < len; i++)
        {

            // Condition to update the answer
            if (str.charAt(i) == '0')
                ans += (i + 1);
        }
        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        String str = "10010";
        int len = str.length();

        System.out.println(countSubStr(str, len));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the count
# of the required substrings
def countSubStr(strr, lenn):

    # To store the final answer
    ans = 0

    # Loop to find the answer
    for i in range(lenn):

        # Condition to update the answer
        if (strr[i] == '0'):
            ans += (i + 1)

    return ans

# Driver code
strr = "10010"
lenn = len(strr)

print(countSubStr(strr, lenn))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the count
    // of the required substrings
    static int countSubStr(string str, int len)
    {
        // To store the final answer
        int ans = 0;

        // Loop to find the answer
        for (int i = 0; i < len; i++)
        {

            // Condition to update the answer
            if (str[i] == '0')
                ans += (i + 1);
        }
        return ans;
    }

    // Driver code
    public static void Main ()
    {
        string str = "10010";
        int len = str.Length;

        Console.WriteLine(countSubStr(str, len));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count
// of the required substrings
function countSubStr(str, len)
{
    // To store the final answer
    var ans = 0;

    // Loop to find the answer
    for (var i = 0; i < len; i++) {

        // Condition to update the answer
        if (str[i] == '0')
            ans += (i + 1);
    }

    return ans;
}

// Driver code
var str = "10010";
var len = str.length;
document.write( countSubStr(str, len));

</script>

Output:

10

Time Complexity: O(N), N = String length

Auxiliary Space: O(1)