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; } |
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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 |
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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 |
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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 |
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Output:
geeks
Time Complexity: O(N * M) where N, M are the lengths of given strings.
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