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Largest sub-string of a binary string divisible by 2

  • Last Updated : 13 May, 2021

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. 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 <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

 


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