Longest Common Prefix using Divide and Conquer Algorithm
Given a set of strings, find the longest common prefix.
Examples:
Input : {“geeksforgeeks”, “geeks”, “geek”, “geezer”} Output : "gee" Input : {"apple", "ape", "april"} Output : "ap"
We have discussed word by word matching and character by character matching algorithms.
In this algorithm, a divide and conquer approach is discussed. We first divide the arrays of string into two parts. Then we do the same for left part and after that for the right part. We will do it until and unless all the strings become of length 1. Now after that, we will start conquering by returning the common prefix of the left and the right strings.
The algorithm will be clear using the below illustration. We consider our strings as – “geeksforgeeks”, “geeks”, “geek”, “geezer”.
Below is the implementation.
C++
// A C++ Program to find the longest common prefix #include<bits/stdc++.h> using namespace std; // A Utility Function to find the common prefix between // strings- str1 and str2 string commonPrefixUtil(string str1, string str2) { string result; int n1 = str1.length(), n2 = str2.length(); for ( int i=0, j=0; i<=n1-1&&j<=n2-1; i++,j++) { if (str1[i] != str2[j]) break ; result.push_back(str1[i]); } return (result); } // A Divide and Conquer based function to find the // longest common prefix. This is similar to the // merge sort technique string commonPrefix(string arr[], int low, int high) { if (low == high) return (arr[low]); if (high > low) { // Same as (low + high)/2, but avoids overflow for // large low and high int mid = low + (high - low) / 2; string str1 = commonPrefix(arr, low, mid); string str2 = commonPrefix(arr, mid+1, high); return (commonPrefixUtil(str1, str2)); } } // Driver program to test above function int main() { string arr[] = { "geeksforgeeks" , "geeks" , "geek" , "geezer" }; int n = sizeof (arr) / sizeof (arr[0]); string ans = commonPrefix(arr, 0, n-1); if (ans.length()) cout << "The longest common prefix is " << ans; else cout << "There is no common prefix" ; return (0); } |
Java
// Java Program to find the longest common prefix class GFG { // A Utility Function to find the common prefix between // strings- str1 and str2 static String commonPrefixUtil(String str1, String str2) { String result = "" ; int n1 = str1.length(), n2 = str2.length(); for ( int i = 0 , j = 0 ; i <= n1 - 1 && j <= n2 - 1 ; i++, j++) { if (str1.charAt(i) != str2.charAt(j)) { break ; } result += str1.charAt(i); } return (result); } // A Divide and Conquer based function to find the // longest common prefix. This is similar to the // merge sort technique static String commonPrefix(String arr[], int low, int high) { if (low == high) { return (arr[low]); } if (high > low) { // Same as (low + high)/2, but avoids overflow for // large low and high int mid = low + (high - low) / 2 ; String str1 = commonPrefix(arr, low, mid); String str2 = commonPrefix(arr, mid + 1 , high); return (commonPrefixUtil(str1, str2)); } return null ; } // Driver program to test above function public static void main(String[] args) { String arr[] = { "geeksforgeeks" , "geeks" , "geek" , "geezer" }; int n = arr.length; String ans = commonPrefix(arr, 0 , n - 1 ); if (ans.length() != 0 ) { System.out.println( "The longest common prefix is " + ans); } else { System.out.println( "There is no common prefix" ); } } } /* This JAVA code is contributed by 29AjayKumar*/ |
Python3
# A Python3 Program to find the longest common prefix # A Utility Function to find the common # prefix between strings- str1 and str2 def commonPrefixUtil(str1, str2): result = "" n1, n2 = len (str1), len (str2) i, j = 0 , 0 while i < = n1 - 1 and j < = n2 - 1 : if str1[i] ! = str2[j]: break result + = str1[i] i, j = i + 1 , j + 1 return result # A Divide and Conquer based function to # find the longest common prefix. This is # similar to the merge sort technique def commonPrefix(arr, low, high): if low = = high: return arr[low] if high > low: # Same as (low + high)/2, but avoids # overflow for large low and high mid = low + (high - low) / / 2 str1 = commonPrefix(arr, low, mid) str2 = commonPrefix(arr, mid + 1 , high) return commonPrefixUtil(str1, str2) # Driver Code if __name__ = = "__main__" : arr = [ "geeksforgeeks" , "geeks" , "geek" , "geezer" ] n = len (arr) ans = commonPrefix(arr, 0 , n - 1 ) if len (ans): print ( "The longest common prefix is" , ans) else : print ( "There is no common prefix" ) # This code is contributed by Rituraj Jain |
C#
// C# Program to find the longest // common prefix using System; class GFG { // A Utility Function to find // the common prefix between // strings- str1 and str2 static string commonPrefixUtil( string str1, string str2) { string result = "" ; int n1 = str1.Length, n2 = str2.Length; for ( int i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) { if (str1[i] != str2[j]) break ; result += str1[i]; } return (result); } // A Divide and Conquer based // function to find the longest // common prefix. This is similar // to the merge sort technique static string commonPrefix( string []arr, int low, int high) { if (low == high) return (arr[low]); if (high > low) { // Same as (low + high)/2, // but avoids overflow for // large low and high int mid = low + (high - low) / 2; string str1 = commonPrefix(arr, low, mid); string str2 = commonPrefix(arr, mid + 1, high); return (commonPrefixUtil(str1, str2)); } return null ; } // Driver Code public static void Main() { String []arr = { "geeksforgeeks" , "geeks" , "geek" , "geezer" }; int n = arr.Length; String ans = commonPrefix(arr, 0, n - 1); if (ans.Length!= 0) { Console.Write( "The longest common " + "prefix is " + ans); } else { Console.Write( "There is no common prefix" ); } } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program to find the // longest common prefix // A Utility Function to find the // common prefix between strings- // str1 and str2 function commonPrefixUtil(str1, str2) { let result = "" ; let n1 = str1.length, n2 = str2.length; for (let i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) { if (str1[i] != str2[j]) { break ; } result += str1[i]; } return (result); } // A Divide and Conquer based function // to find the longest common prefix. This // is similar to the merge sort technique function commonPrefix(arr, low, high) { if (low == high) { return (arr[low]); } if (high > low) { // Same as (low + high)/2, but avoids // overflow for large low and high let mid = low + Math.floor((high - low) / 2); let str1 = commonPrefix(arr, low, mid); let str2 = commonPrefix(arr, mid + 1, high); return (commonPrefixUtil(str1, str2)); } return null ; } // Driver code let arr = [ "geeksforgeeks" , "geeks" , "geek" , "geezer" ]; let n = arr.length; let ans = commonPrefix(arr, 0, n - 1); if (ans.length != 0) { document.write( "The longest common prefix is " + ans); } else { document.write( "There is no common prefix" ); } // This code is contributed by rag2127 </script> |
The longest common prefix is gee
Time Complexity: Since we are iterating through all the characters of all the strings, so we can say that the time complexity is O(N M) where,
N = Number of strings M = Length of the largest string string
Auxiliary Space: To store the longest prefix string we are allocating space which is O(M Log N).
This article is contributed by Rachit Belwariar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Login to comment...