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(N^{3}).

**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 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

`<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)

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