Skip to content
Related Articles
Number of sub-strings in a given binary string divisible by 2
• Last Updated : 13 May, 2021

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:
“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 ``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 substrrings``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

 ``
Output:
`10`

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

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 industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up