Skip to content
Related Articles

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.

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 <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 
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
Recommended Articles
Page :