Longest Common Prefix using Character by Character Matching
Given a set of strings, find the longest common prefix.
Input : {“w3wiki”, “Beginner”, “geek”, “geezer”} Output : "gee" Input : {"apple", "ape", "april"} Output : "ap"
We have discussed word by word matching algorithm in previous post.
In this algorithm, instead of going through the strings one by one, we will go through the characters one by one.
We consider our strings as – “w3wiki”, “Beginner”, “geek”, “geezer”.
Below is the implementation of this approach.
C++
// A C++ Program to find the longest common prefix #include<bits/stdc++.h> using namespace std; // A Function to find the string having the minimum // length and returns that length int findMinLength(string arr[], int n) { int min = arr[0].length(); for ( int i=1; i<n; i++) if (arr[i].length() < min) min = arr[i].length(); return (min); } // A Function that returns the longest common prefix // from the array of strings string commonPrefix(string arr[], int n) { int minlen = findMinLength(arr, n); string result; // Our resultant string char current; // The current character for ( int i=0; i<minlen; i++) { // Current character (must be same // in all strings to be a part of // result) current = arr[0][i]; for ( int j=1 ; j<n; j++) if (arr[j][i] != current) return result; // Append to result result.push_back(current); } return (result); } // Driver program to test above function int main() { string arr[] = { "w3wiki" , "Beginner" , "geek" , "geezer" }; int n = sizeof (arr) / sizeof (arr[0]); string ans = commonPrefix (arr, n); if (ans.length()) cout << "The longest common prefix is " << ans; else cout << "There is no common prefix" ; return (0); } |
Java
// A Java Program to find the longest common prefix class GFG { // A Function to find the string having the minimum // length and returns that length static int findMinLength(String arr[], int n) { int min = arr[ 0 ].length(); for ( int i = 1 ; i < n; i++) { if (arr[i].length() < min) { min = arr[i].length(); } } return (min); } // A Function that returns the longest common prefix // from the array of strings static String commonPrefix(String arr[], int n) { int minlen = findMinLength(arr, n); String result = "" ; // Our resultant string char current; // The current character for ( int i = 0 ; i < minlen; i++) { // Current character (must be same // in all strings to be a part of // result) current = arr[ 0 ].charAt(i); for ( int j = 1 ; j < n; j++) { if (arr[j].charAt(i) != current) { return result; } } // Append to result result += (current); } return (result); } // Driver program to test above function public static void main(String[] args) { String arr[] = { "w3wiki" , "Beginner" , "geek" , "geezer" }; int n = arr.length; String ans = commonPrefix(arr, n); if (ans.length() > 0 ) { System.out.println( "The longest common prefix is " + ans); } else { System.out.println( "There is no common prefix" ); } } } // This code contributed by Rajput-Ji |
Python 3
# Python 3 Program to find the longest common prefix # A Function to find the string having the minimum # length and returns that length def findMinLength(arr, n): min = len (arr[ 0 ]) for i in range ( 1 ,n): if ( len (arr[i])< min ): min = len (arr[i]) return ( min ) # A Function that returns the longest common prefix # from the array of strings def commonPrefix(arr, n): minlen = findMinLength(arr, n) result = "" for i in range (minlen): # Current character (must be same # in all strings to be a part of # result) current = arr[ 0 ][i] for j in range ( 1 ,n): if (arr[j][i] ! = current): return result # Append to result result = result + current return (result) # Driver program to test above function if __name__ = = "__main__" : arr = [ "w3wiki" , "Beginner" , "geek" , "geezer" ] n = len (arr) ans = commonPrefix (arr, n) if ( len (ans)): print ( "The longest common prefix is " ,ans) else : print ( "There is no common prefix" ) |
C#
// A C# Program to find the longest common prefix using System; class GFG { // A Function to find the string having the minimum // length and returns that length static int findMinLength(String []arr, int n) { int min = arr[0].Length; for ( int i = 1; i < n; i++) { if (arr[i].Length < min) { min = arr[i].Length; } } return (min); } // A Function that returns the longest common prefix // from the array of strings static String commonPrefix(String []arr, int n) { int minlen = findMinLength(arr, n); String result = "" ; // Our resultant string char current; // The current character for ( int i = 0; i < minlen; i++) { // Current character (must be same // in all strings to be a part of // result) current = arr[0][i]; for ( int j = 1; j < n; j++) { if (arr[j][i] != current) { return result; } } // Append to result result += (current); } return (result); } // Driver code public static void Main(String[] args) { String []arr = { "w3wiki" , "Beginner" , "geek" , "geezer" }; int n = arr.Length; String ans = commonPrefix(arr, n); if (ans.Length > 0) { Console.WriteLine( "The longest common prefix is " + ans); } else { Console.WriteLine( "There is no common prefix" ); } } } /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // A Javascript Program to find the longest common prefix // A Function to find the string having the minimum // length and returns that length function findMinLength(arr,n) { let min = arr[0].length; for (let i = 1; i < n; i++) { if (arr[i].length < min) { min = arr[i].length; } } return (min); } // A Function that returns the longest common prefix // from the array of strings function commonPrefix(arr,n) { let minlen = findMinLength(arr, n); let result = "" ; // Our resultant string let current; // The current character for (let i = 0; i < minlen; i++) { // Current character (must be same // in all strings to be a part of // result) current = arr[0][i]; for (let j = 1; j < n; j++) { if (arr[j][i] != current) { return result; } } // Append to result result += (current); } return (result); } // Driver program to test above function let arr=[ "w3wiki" , "Beginner" , "geek" , "geezer" ] let n = arr.length; let ans = commonPrefix(arr, n); 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 avanitrachhadiya2155 </script> |
The longest common prefix is gee
How is this algorithm better than the “Word by Word Matching” algorithm ?-
In Set 1 we discussed about the “Word by Word Matching” Algorithm.
Suppose you have the input strings as- “w3wiki”, “Beginner”, “geek”, “geezer”, “x”.
Now there is no common prefix string of the above strings. By the “Word by Word Matching” algorithm discussed in Set 1, we come to the conclusion that there is no common prefix string by traversing all the strings. But if we use this algorithm, then in the first iteration itself we will come to know that there is no common prefix string, as we don’t go further to look for the second character of each strings.
This algorithm has a huge advantage when there are too many strings.
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 Smallest string
Auxiliary Space : To store the longest prefix string we are allocating space which is O(M).
Method 3(using some in-built C++ STL function):
In this approach first of all we will find the string with smallest length. Then we will search in all other strings if this smallest string is present in them as a prefix or not. If in all strings this small string is present is present then we will print this string else we will keep reducing the length of the smallest string by one until its length becomes zero.
Below is the implementation of the above Approach:
Implementation:
C++
// C++ program to find the longest common // prefix. #include <bits/stdc++.h> using namespace std; // function to find the smallest string // among all string int shortest_string(string s[], int n) { int minlength = INT_MAX, min_index; for ( int i = 0; i < n; i++) { if (s[i].length() < minlength) { minlength = s[i].length(); min_index = i; } } return min_index; } // function to find longest common // prefix among all strings. string findprefix(string s[], int n) { // index of the smallest string int shortest_string_index = shortest_string(s, n); while (s[shortest_string_index].length() > 0) { int count = 0; for ( int i = 0; i < n; i++) { // checking whether all strings have prefix // which is equal to smallest string if (s[i].find(s[shortest_string_index]) == 0) { count++; } } // checking that all the string's // prefix is equal to smallest string // or not. if (count == n) { cout << "longest common prefix is: " << endl; return s[shortest_string_index]; break ; } // deleting the last character // of the smallest string. s[shortest_string_index].pop_back(); } return "no common prefix among all strings" ; } // driver code int main() { string s[] = { "w3wiki" , "Beginner" , "geek" , "geezer" }; int n = sizeof (s) / sizeof (s[0]); // function call cout << findprefix(s, n); return 0; } // this code is contributed by Machhaliya Mohammad. |
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { // function to find the smallest string // among all string static int shortest_string(String s[], int n) { int minlength = Integer.MAX_VALUE, min_index = - 1 ; for ( int i = 0 ; i < n; i++) { if (s[i].length() < minlength) { minlength = s[i].length(); min_index = i; } } return min_index; } // function to find longest common // prefix among all strings. static String findprefix(String s[], int n) { // index of the smallest string int shortest_string_index = shortest_string(s, n); while (s[shortest_string_index].length() > 0 ) { int count = 0 ; for ( int i = 0 ; i < n; i++) { // checking whether all strings have prefix // which is equal to smallest string if (s[i].indexOf(s[shortest_string_index]) == 0 ) { count++; } } // checking that all the string's // prefix is equal to smallest string // or not. if (count == n) { System.out.println( "longest common prefix is:" ); return s[shortest_string_index]; } // deleting the last character // of the smallest string. s[shortest_string_index] = s[shortest_string_index].substring( 0 ,s[shortest_string_index].length()- 1 ); } return "no common prefix among all strings" ; } public static void main (String[] args) { String s[] = { "w3wiki" , "Beginner" , "geek" , "geezer" }; int n = s.length; // function call System.out.println(findprefix(s, n)); } } // This code is contributed by aadityaburujwale. |
Python3
# Python program to find the longest common # prefix. import sys # function to find the smallest string # among all string def shortest_string(s, n): minlength = sys.maxsize min_index = 0 for i in range (n): if len (s[i]) < minlength: minlength = len (s[i]) min_index = i return min_index # function to find longest common # prefix among all strings. def findprefix(s, n): # index of the smallest string shortest_string_index = shortest_string(s, n) while len (s[shortest_string_index]) > 0 : count = 0 for i in range (n): # checking whether all strings have prefix # which is equal to smallest string if s[i].find(s[shortest_string_index]) = = 0 : count + = 1 # checking that all the string's # prefix is equal to smallest string # or not. if count = = n: print ( "longest common prefix is: " ) return s[shortest_string_index] break # deleting the last character # of the smallest string. s[shortest_string_index] = s[shortest_string_index][: - 1 ] return "no common prefix among all strings" # driver code s = [ "w3wiki" , "Beginner" , "geek" , "geezer" ] n = len (s) # function call print (findprefix(s, n)) # This code is contributed by akashish__ |
C#
// Include namespace system using System; public class GFG { // function to find the smallest string // among all string public static int shortest_string(String[] s, int n) { var minlength = int .MaxValue; var min_index = -1; for ( int i = 0; i < n; i++) { if (s[i].Length < minlength) { minlength = s[i].Length; min_index = i; } } return min_index; } // function to find longest common // prefix among all strings. public static String findprefix(String[] s, int n) { // index of the smallest string var shortest_string_index = GFG.shortest_string(s, n); while (s[shortest_string_index].Length > 0) { var count = 0; for ( int i = 0; i < n; i++) { // checking whether all strings have prefix // which is equal to smallest string if (s[i].IndexOf(s[shortest_string_index]) == 0) { count++; } } // checking that all the string's // prefix is equal to smallest string // or not. if (count == n) { Console.WriteLine( "longest common prefix is:" ); return s[shortest_string_index]; } // deleting the last character // of the smallest string. s[shortest_string_index] = s[shortest_string_index].Substring(0,s[shortest_string_index].Length - 1-0); } return "no common prefix among all strings" ; } public static void Main(String[] args) { String[] s = { "w3wiki" , "Beginner" , "geek" , "geezer" }; var n = s.Length; // function call Console.WriteLine(GFG.findprefix(s, n)); } } // This code is contributed by sourabhdalal0001. |
Javascript
// JavaScript program to find the longest common // prefix. function shortest_string(s, n) { let minlength = Number.MAX_SAFE_INTEGER; let min_index; for (let i = 0; i < n; i++) { if (s[i].length < minlength) { minlength = s[i].length; min_index = i; } } return min_index; } function findprefix(s, n) { // index of the smallest string let shortest_string_index = shortest_string(s, n); while (s[shortest_string_index].length > 0) { let count = 0; for (let i = 0; i < n; i++) { // checking whether all strings have prefix // which is equal to smallest string if (s[i].indexOf(s[shortest_string_index]) === 0) { count++; } } // checking that all the string's // prefix is equal to smallest string // or not. if (count === n) { console.log( "longest common prefix is: " ); return s[shortest_string_index]; break ; } // deleting the last character // of the smallest string. s[shortest_string_index] = s[shortest_string_index].slice(0, -1); } return "no common prefix among all strings" ; } // driver code let s = [ "w3wiki" , "Beginner" , "geek" , "geezer" ]; let n = s.length; // function call console.log(findprefix(s, n)); // This code is contributed by akashish__ |
longest common prefix is: gee
Time complexity: O(n*m) where n is total number of strings and m is length of the smallest string among all strings.
Auxiliary space: O(1)