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# 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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