Given an array of strings arr[] of size N, the task is to print all the distinct strings present in the given array.
Examples:
Input: arr[] = { “Geeks”, “For”, “Geeks”, “Code”, “Coder” }
Output: Coder Code Geeks For
Explanation: Since all the strings in the array are distinct, the required output is Coder Code Geeks For.Input: arr[] = { “Good”, “God”, “Good”, “God”, “god” }
Output: god Good God
Naive Approach: The simplest approach to solve this problem is to sort the array based on the lexicographical order of the strings. Traverse the array and check if the current string of the array is equal to the previously traversed string or not. If found to be false, then print the current string.
Time Complexity: O(N * M * log(N)), where M is the length of the longest string.
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Hashing. Follow the steps below to solve the problem:
- Initialize a Set, say DistString, to store the distinct strings from the given array.
- Traverse the array and insert array elements into DistString.
- Finally, print all strings from DistString.
Below is the implementation of the above approach.
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the distinct strings// from the given arrayvoid findDisStr(vector<string>& arr, int N){ // Stores distinct strings // from the given array unordered_set<string> DistString; // Traverse the array for (int i = 0; i < N; i++) { // If current string not // present into the set if (!DistString.count(arr[i])) { // Insert current string // into the set DistString.insert(arr[i]); } } // Traverse the set DistString for (auto String : DistString) { // Print distinct string cout << String << " "; }}// Driver Codeint main(){ vector<string> arr = { "Geeks", "For", "Geeks", "Code", "Coder" }; // Stores length of the array int N = arr.size(); findDisStr(arr, N); return 0;} |
Java
// Java program to implement// the above approachimport java.io.*;import java.util.*;class GFG{// Function to find the distinct strings// from the given arraystatic void findDisStr(List<String> arr, int N){ // Stores distinct strings // from the given array Set<String> DistString = new HashSet<String>(); // Traverse the array for(int i = 0; i < N; i++) { // If current string not // present into the set if (!DistString.contains(arr.get(i))) { // Insert current string // into the set DistString.add(arr.get(i)); } } // Traverse the set DistString for(String string : DistString) { // Print distinct string System.out.print(string + " "); }}// Driver codepublic static void main(String[] args){ List<String> arr = Arrays.asList(new String[]{ "Geeks", "For", "Geeks", "Code", "Coder" }); // Stores length of the array int N = arr.size(); findDisStr(arr, N);}}// This code is contributed by jithin |
Python3
# Python3 program to implement# the above approach# Function to find the distinct # strings from the given arraydef findDisStr(arr, N): # Stores distinct strings # from the given array DistString = set() # Traverse the array for i in range(N): # If current string not # present into the set if (arr[i] not in DistString): # Insert current string # into the set DistString.add(arr[i]) # Traverse the set DistString for string in DistString: # Print distinct string print(string, end = " ") # Driver Codeif __name__ == "__main__": arr = [ "Geeks", "For", "Geeks", "Code", "Coder" ] # Stores length of the array N = len(arr) findDisStr(arr, N) # This code is contributed by chitranayal |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the distinct strings// from the given arraystatic void findDisStr(List<string> arr, int N){ // Stores distinct strings // from the given array HashSet<string> DistString = new HashSet<string>(); // Traverse the array for(int i = 0; i < N; i++) { // If current string not // present into the set if (!DistString.Contains(arr[i])) { // Insert current string // into the set DistString.Add(arr[i]); } } // Traverse the set DistString foreach(string a in DistString) { Console.Write(a +" "); } }// Driver codepublic static void Main(String[] args){ List<String> arr = new List<string>(new []{ "Geeks", "For", "Geeks", "Code", "Coder" }); // Stores length of the array int N = arr.Count; findDisStr(arr, N);}}// This code is contributed by jana_sayantan |
Javascript
<script>// JavaScript program to implement// the above approach// Function to find the distinct strings// from the given arrayfunction findDisStr(arr, N) { // Stores distinct strings // from the given array let DistString = new Set(); // Traverse the array for (let i = N - 1; i >= 0; i--) { // If current string not // present into the set if (!DistString.has(arr[i])) { // Insert current string // into the set DistString.add(arr[i]); } } for (let String of DistString) { // Print distinct string document.write(String + " "); }}// Driver Codelet arr = ["Geeks", "For", "Geeks", "Code", "Coder"];// Stores length of the arraylet N = arr.length;findDisStr(arr, N);</script> |
Output:
Coder Code Geeks For
Time Complexity: O(N * M), where M is the length of the longest string.
Auxiliary Space: O(N * M)
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