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++ 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 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 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# 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 

<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>   

10

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

Auxiliary Space: O(1)