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

• Last Updated : 25 Aug, 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 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

 ``

Output:

`10`

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

Auxiliary Space: O(1)

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