Problem Statement: Given a set of strings, find the longest common prefix.
Examples:
Input: {“neveropen”, “neveropen”, “geek”, “geezer”}
Output: “gee”Input: {“apple”, “ape”, “april”}
Output: “ap”
The longest common prefix for an array of strings is the common prefix between 2 most dissimilar strings. For example, in the given array {“apple”, “ape”, “zebra”}, there is no common prefix because the 2 most dissimilar strings of the array “ape” and “zebra” do not share any starting characters.
We have discussed five different approaches below posts.
In this post, a new method based on sorting is discussed. The idea is to sort the array of strings and find the common prefix of the first and last string of the sorted array.
C++
// C++ program to find longest common prefix // of given array of words.#include<iostream>#include<algorithm>using namespace std;// Function to find the longest common prefixstring longestCommonPrefix(string ar[], int n){ // If size is 0, return empty string if (n == 0) return ""; //If size is 1 then just return that character if (n == 1) return ar[0]; // Sort the given array sort(ar, ar + n); // Find the minimum length from // first and last string int en = min(ar[0].size(), ar[n - 1].size()); // Now the common prefix in first and // last string is the longest common prefix string first = ar[0], last = ar[n - 1]; int i = 0; while (i < en && first[i] == last[i]) i++; string pre = first.substr(0, i); return pre;}// Driver Codeint main(){ string ar[] = {"neveropen", "neveropen", "geek", "geezer"}; int n = sizeof(ar) / sizeof(ar[0]); cout << "The longest common prefix is: " << longestCommonPrefix(ar, n); return 0;}// This code is contributed by jrolofmeister |
Java
// Java program to find longest common prefix of// given array of words.import java.util.*;public class GFG{ public String longestCommonPrefix(String[] a) { int size = a.length; /* if size is 0, return empty string */ if (size == 0) return ""; if (size == 1) return a[0]; /* sort the array of strings */ Arrays.sort(a); /* find the minimum length from first and last string */ int end = Math.min(a[0].length(), a[size-1].length()); /* find the common prefix between the first and last string */ int i = 0; while (i < end && a[0].charAt(i) == a[size-1].charAt(i) ) i++; String pre = a[0].substring(0, i); return pre; } /* Driver Function to test other function */ public static void main(String[] args) { GFG gfg = new GFG(); String[] input = {"neveropen", "neveropen", "geek", "geezer"}; System.out.println( "The longest Common Prefix is : " + gfg.longestCommonPrefix(input)); }} |
Python 3
# Python 3 program to find longest # common prefix of given array of words.def longestCommonPrefix( a): size = len(a) # if size is 0, return empty string if (size == 0): return "" if (size == 1): return a[0] # sort the array of strings a.sort() # find the minimum length from # first and last string end = min(len(a[0]), len(a[size - 1])) # find the common prefix between # the first and last string i = 0 while (i < end and a[0][i] == a[size - 1][i]): i += 1 pre = a[0][0: i] return pre# Driver Codeif __name__ == "__main__": input = ["neveropen", "neveropen", "geek", "geezer"] print("The longest Common Prefix is :" , longestCommonPrefix(input))# This code is contributed by ita_c |
C#
// C# program to find longest common prefix of// given array of words.using System; public class GFG { static string longestCommonPrefix(String[] a) { int size = a.Length; /* if size is 0, return empty string */ if (size == 0) return ""; if (size == 1) return a[0]; /* sort the array of strings */ Array.Sort(a); /* find the minimum length from first and last string */ int end = Math.Min(a[0].Length, a[size-1].Length); /* find the common prefix between the first and last string */ int i = 0; while (i < end && a[0][i] == a[size-1][i] ) i++; string pre = a[0].Substring(0, i); return pre; } /* Driver Function to test other function */ public static void Main() { string[] input = {"neveropen", "neveropen", "geek", "geezer"}; Console.WriteLine( "The longest Common" + " Prefix is : " + longestCommonPrefix(input)); }}// This code is contributed by Sam007. |
Javascript
<script>// Javascript program to find longest common prefix of// given array of words. function longestCommonPrefix(a) { let size = a.length; /* if size is 0, return empty string */ if (size == 0) return ""; if (size == 1) return a[0]; /* sort the array of strings */ a.sort(); /* find the minimum length from first and last string */ let end = Math.min(a[0].length, a[size-1].length); /* find the common prefix between the first and last string */ let i = 0; while (i < end && a[0][i] == a[size-1][i] ) i++; let pre = a[0].substring(0, i); return pre; } /* Driver Function to test other function */ let input=["neveropen", "neveropen", "geek", "geezer"]; document.write( "The longest Common Prefix is : " + longestCommonPrefix(input)); // This code is contributed by rag2127</script> |
Output
The longest common prefix is: gee
Time Complexity: O(MAX * n * log n ) where n is the number of strings in the array and MAX is the maximum number of characters in any string. Please note that comparison of two strings would take at most O(MAX) time, and for sorting n strings, we would need O(MAX * n * log n ) time.
Auxiliary Space: O(1)
Approach (2) :
The Idea is to find unique prefix without sorting the string array. Extract substring by comparing current substring with the previous substring and update it’s result. For large string it work much faster.
C++
#include <bits/stdc++.h>using namespace std;string longestCommonPrefix(string arr[], int n){ // take temp_prefix string and assign first element of // arr : a. string result = arr[0]; int len = result.length(); // Iterate for rest of element in string. for (int i = 1; i < n; i++) { // .find() return index of that substring from // string. while (arr[i].find(result) != 0) { // update matched substring prefx result = result.substr(0, len - 1); len--; // check for empty case. direct return if true.. if (result.empty()) { return "-1"; } } } return result;}/* Driver Function to test other function */int main(){ string input[] = { "neveropen", "neveropen", "geek", "geezer" }; int n = sizeof(input) / sizeof(input[0]); cout << "The longest Common Prefix is : " << longestCommonPrefix(input, n); return 0;} |
Java
// Java program to find longest common prefix of// given array of words.import java.util.*;public class GFG{ public String longestCommonPrefix(String[] arr) { int n = arr.length; // take temp_prefix string and assign first element of arr : a. String result = arr[0]; // Iterate for rest of element in string. for(int i = 1; i < n; i++){ // .indexOf() return index of that substring from string. while(arr[i].indexOf(result) != 0){ // update matched substring prefx result = result.substring(0, result.length()-1); // check for empty case. direct return if true.. if(result.isEmpty()){ return "-1"; } } } return result; } /* Driver Function to test other function */ public static void main(String[] args) { GFG gfg = new GFG(); String[] input = {"neveropen", "neveropen", "geek", "geezer"}; System.out.println( "The longest Common Prefix is : " + gfg.longestCommonPrefix(input)); }}// this code is contributed by shantanu_mourya |
Python3
def longestCommonPrefix(arr): # Assign the first element of arr to result result = arr[0] length = len(result) # Iterate for the rest of the elements in the array for i in range(1, len(arr)): # Find the index of result in the current string while arr[i].find(result) != 0: # Update the matched substring prefix result = result[:length - 1] length -= 1 # Check for an empty case and return if true if not result: return "-1" return result# Driver code to test the functionif __name__ == "__main__": input_list = ["neveropen", "neveropen", "geek", "geezer"] print("The longest Common Prefix is:", longestCommonPrefix(input_list)) |
C#
using System;class Program{ static string LongestCommonPrefix(string[] arr) { // Assign the first element of arr to result string result = arr[0]; int length = result.Length; // Iterate for the rest of the elements in the array for (int i = 1; i < arr.Length; i++) { // Find the index of result in the current string while (arr[i].IndexOf(result) != 0) { // Update the matched substring prefix result = result.Substring(0, length - 1); length--; // Check for an empty case and return if true if (result.Length == 0) { return "-1"; } } } return result; } static void Main() { string[] input = { "neveropen", "neveropen", "geek", "geezer" }; Console.WriteLine("The longest Common Prefix is: " + LongestCommonPrefix(input)); }} |
Javascript
function longestCommonPrefix(arr) { // Assign the first element of arr to result let result = arr[0]; let length = result.length; // Iterate for the rest of the elements in the array for (let i = 1; i < arr.length; i++) { // Find the index of result in the current string while (arr[i].indexOf(result) !== 0) { // Update the matched substring prefix result = result.substring(0, length - 1); length--; // Check for an empty case and return if true if (result === '') { return '-1'; } } } return result;}// Driver code to test the functionconst input = ["neveropen", "neveropen", "geek", "geezer"];console.log("The longest Common Prefix is:", longestCommonPrefix(input)); |
Output
The longest Common Prefix is : gee
Time Complexity : O ( m * N). where m is the length of the prefix and n is the number of strings in the input array.
This article is contributed by Saloni Baweja. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
