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 <bits/stdc++.h>
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

<script>

// Javascript implementation of the approach

// Function to return the largest
// substring divisible by 2
function largestSubStr(s)
{
    // While the last character of
    // the string is '1', pop it
    while (s.length && 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
var s = "11001";
document.write( largestSubStr(s));

</script>

Output:

1100

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