Related Articles
Largest sub-string of a binary string divisible by 2
• Last Updated : 03 Jan, 2020

Given a binary string str of length N, the task is to find the longest sub-string divisible by 2. If no such sub-string exists then print -1.

Examples:

Input: str = “11100011”
Output: 111000
Largest sub-string divisible by 2 is “111000”.

Input: str = “1111”
Output: -1
There is no sub-string of the given string
which is divisible by 2.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: A naive approach will be to generate all such sub-strings and check if they are divisible by 2. The time complexity of this approach will be O(N3).

Better approach: A straight forward approach will be to remove characters from the end of the string while the last character is 1. The moment a 0 is encountered, the current string will be divisible by 2 as it ends at a 0. The time complexity of this approach will be O(N).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the largest``// substring divisible by 2``string largestSubStr(string s)``{``    ``// While the last character of``    ``// the string is '1', pop it``    ``while` `(s.size() and s[s.size() - 1] == ``'1'``)``        ``s.pop_back();`` ` `    ``// If the original string had no '0'``    ``if` `(s.size() == 0)``        ``return` `"-1"``;``    ``else``        ``return` `s;``}`` ` `// Driver code``int` `main()``{``    ``string s = ``"11001"``;`` ` `    ``cout << largestSubStr(s);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach ``class` `GFG``{``     ` `    ``// Function to return the largest ``    ``// substring divisible by 2 ``    ``static` `String largestSubStr(String s) ``    ``{ ``        ``// While the last character of ``        ``// the string is '1', pop it ``        ``while` `(s.length() != ``0` `&& ``               ``s.charAt(s.length() - ``1``) == ``'1'``) ``            ``s = s.substring(``0``, s.length() - ``1``); ``     ` `        ``// If the original string had no '0' ``        ``if` `(s.length() == ``0``) ``            ``return` `"-1"``; ``        ``else``            ``return` `s; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args)``    ``{ ``        ``String s = ``"11001"``; ``     ` `        ``System.out.println(largestSubStr(s)); ``    ``} ``}`` ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach `` ` `# Function to return the largest ``# substring divisible by 2 ``def` `largestSubStr(s) : `` ` `    ``# While the last character of ``    ``# the string is '1', pop it ``    ``while` `(``len``(s) ``and` `s[``len``(s) ``-` `1``] ``=``=` `'1'``) :``        ``s ``=` `s[:``len``(s) ``-` `1``]; `` ` `    ``# If the original string had no '0' ``    ``if` `(``len``(s) ``=``=` `0``) :``        ``return` `"-1"``; ``    ``else` `:``        ``return` `s; `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `:`` ` `    ``s ``=` `"11001"``; `` ` `    ``print``(largestSubStr(s)); `` ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach ``using` `System;`` ` `class` `GFG ``{ ``     ` `    ``// Function to return the largest ``    ``// substring divisible by 2 ``    ``static` `string` `largestSubStr(``string` `s) ``    ``{ ``        ``// While the last character of ``        ``// the string is '1', pop it ``        ``while` `(s.Length != 0 && ``               ``s[s.Length - 1] == ``'1'``) ``            ``s = s.Substring(0, s.Length - 1); ``     ` `        ``// If the original string had no '0' ``        ``if` `(s.Length == 0) ``            ``return` `"-1"``; ``        ``else``            ``return` `s; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `Main () ``    ``{ ``        ``string` `s = ``"11001"``; ``     ` `        ``Console.WriteLine(largestSubStr(s)); ``    ``} ``} `` ` `// This code is contributed by AnkitRai01 `
Output:
```1100
```

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 and Geeks Classes Live USA

My Personal Notes arrow_drop_up