Largest substring of str2 which is a prefix of str1

Given two string str1 and str2, the task is to find the longest prefix of str1 which is present as a substring of the string str2. Print the prefix if possible else print -1.
Examples:

Input: str1 = “geeksfor”, str2 = “forgeeks” 
Output: geeks 
All the prefixes of str1 which are present in str2 
are “g”, “ge”, “gee”, “geek” and “geeks”.
Input: str1 = “abc”, str2 = “def” 
Output: -1 

Approach: Check whether str1 is present as a substring in str2. If yes then str1 is the required string else remove the last character from str1 and repeat these steps until either the string str1 becomes empty or the required string is found.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the largest substring
// in str2 which is a prefix of str1
string findPrefix(string str1, string str2)
{

    // To store the index in str2 which
    // matches the prefix in str1
    int pos = -1;

    // While there are characters left in str1
    while (!str1.empty()) {

        // If the prefix is not found in str2
        if (str2.find(str1) == string::npos)

            // Remove the last character
            str1.pop_back();
        else {

            // Prefix found
            pos = str2.find(str1);
            break;
        }
    }

    // No substring found in str2 that
    // matches the prefix of str1
    if (pos == -1)
        return "-1";

    return str1;
}

// Driver code
int main()
{
    string str1 = "geeksfor";
    string str2 = "forgeeks";

    cout << findPrefix(str1, str2);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the largest substring
// in str2 which is a prefix of str1
static String findPrefix(String str1,
                         String str2)
{

    // To store the index in str2 which
    // matches the prefix in str1
    boolean pos = false;

    // While there are characters left in str1
    while (str1.length() > 0)
    {

        // If the prefix is not found in str2
        if (!str2.contains(str1))

            // Remove the last character
            str1 = str1.substring(0, str1.length() - 1);
        else
        {

            // Prefix found
            pos = str2.contains(str1);
            break;
        }
    }

    // No substring found in str2 that
    // matches the prefix of str1
    if (pos == false)
        return "-1";

    return str1;
}

// Driver code
public static void main(String[] args)
{
    String str1 = "geeksfor";
    String str2 = "forgeeks";

    System.out.println(findPrefix(str1, str2));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
import operator

# Function to return the largest substring
# in str2 which is a prefix of str1
def findPrefix(str1, str2):

    # To store the index in str2 which
    # matches the prefix in str1
    pos = False;

    # While there are characters left in str1
    while (len(str1) != 0):

        # If the prefix is not found in str2
        if operator.contains(str2, str1) != True:

            # Remove the last character
            str1 = str1[0: len(str1) - 1];
        else:

            # Prefix found
            pos = operator.contains(str2, str1);
            break;

    # No substring found in str2 that
    # matches the prefix of str1
    if (pos == False):
        return "-1";

    return str1;

# Driver code
if __name__ == '__main__':
    str1 = "geeksfor";
    str2 = "forgeeks";

    print(findPrefix(str1, str2));

# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the largest substring
// in str2 which is a prefix of str1
static String findPrefix(String str1,
                         String str2)
{

    // To store the index in str2 which
    // matches the prefix in str1
    bool pos = false;

    // While there are characters left in str1
    while (str1.Length > 0)
    {

        // If the prefix is not found in str2
        if (!str2.Contains(str1))

            // Remove the last character
            str1 = str1.Substring(0, str1.Length - 1);
        else
        {

            // Prefix found
            pos = str2.Contains(str1);
            break;
        }
    }

    // No substring found in str2 that
    // matches the prefix of str1
    if (pos == false)
        return "-1";

    return str1;
}

// Driver code
public static void Main(String[] args)
{
    String str1 = "geeksfor";
    String str2 = "forgeeks";

    Console.WriteLine(findPrefix(str1, str2));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript implementation of the approach

// Function to return the largest substring
// in str2 which is a prefix of str1
function findPrefix(str1, str2)
{

    // To store the index in str2 which
    // matches the prefix in str1
    var pos = -1;

    // While there are characters left in str1
    while (str1.length!=0) {

        // If the prefix is not found in str2
        if (!str2.includes(str1))

            // Remove the last character
            str1 = str1.substring(0,str1.length-1)
        else {

            // Prefix found
            pos = str2.includes(str1);
            break;
        }
    }

    // No substring found in str2 that
    // matches the prefix of str1
    if (pos == -1)
        return "-1";

    return str1;
}

// Driver code
var str1 = "geeksfor";
var str2 = "forgeeks";
document.write( findPrefix(str1, str2));

// This code is contributed by rutvik_56.
</script>

Output:

geeks

Time Complexity: O(N * M) where N, M are the lengths of given strings.