# Largest sub-string of a binary string divisible by 2

Given 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.

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 straightforward 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

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 `

## Javascript

 ``

Output:

`1100`

Time Complexity: O(n), where n is the length of the string s
Auxiliary Space: O(1)

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