Largest sub-string of a binary string divisible by 2

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output:

1100

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Improved By : AnkitRai01