Given an array arr[] of strings containing the names of employees in a company. Assuming that the names are being entered into a system one after another, the task is to check whether the current name is entered for the first time or not.
Examples:
Input: arr[] = {“neveropen”, “for”, “neveropen”}
Output:
No
No
YesInput: arr[] = {“abc”, “aaa”, “cba”}
Output:
No
No
No
Approach: Create an unordered_set to store the names of the employees and start traversing the array, if the current name is already present in the set then print Yes else print No and insert it into the set.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to insert the names// and check whether they appear// for the first timevoid insertNames(string arr[], int n){ // To store the names // of the employees unordered_set<string> set; for (int i = 0; i < n; i++) { // If current name is appearing // for the first time if (set.find(arr[i]) == set.end()) { cout << "No\n"; set.insert(arr[i]); } else { cout << "Yes\n"; } }}// Driver codeint main(){ string arr[] = { "neveropen", "for", "neveropen" }; int n = sizeof(arr) / sizeof(string); insertNames(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to insert the names// and check whether they appear// for the first timestatic void insertNames(String arr[], int n){ // To store the names // of the employees HashSet<String> set = new HashSet<String>(); for (int i = 0; i < n; i++) { // If current name is appearing // for the first time if (!set.contains(arr[i])) { System.out.print("No\n"); set.add(arr[i]); } else { System.out.print("Yes\n"); } }}// Driver codepublic static void main(String[] args){ String arr[] = { "neveropen", "for", "neveropen" }; int n = arr.length; insertNames(arr, n);}}// This code contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach# Function to insert the names # and check whether they appear # for the first time def insertNames(arr, n) : # To store the names # of the employees string = set(); for i in range(n) : # If current name is appearing # for the first time if arr[i] not in string : print("No"); string.add(arr[i]); else : print("Yes"); # Driver code if __name__ == "__main__" : arr = [ "neveropen", "for", "neveropen" ]; n = len(arr); insertNames(arr, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {// Function to insert the names// and check whether they appear// for the first timestatic void insertNames(String []arr, int n){ // To store the names // of the employees HashSet<String> set = new HashSet<String>(); for (int i = 0; i < n; i++) { // If current name is appearing // for the first time if (!set.Contains(arr[i])) { Console.Write("No\n"); set.Add(arr[i]); } else { Console.Write("Yes\n"); } }}// Driver codepublic static void Main(String[] args){ String []arr = { "neveropen", "for", "neveropen" }; int n = arr.Length; insertNames(arr, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach // Function to insert the names// and check whether they appear// for the first timefunction insertNames(arr, n){ // To store the names // of the employees let set = new Set(); for (let i = 0; i < n; i++) { // If current name is appearing // for the first time if (!set.has(arr[i])) { document.write("No" + "<br/>"); set.add(arr[i]); } else { document.write("Yes" + "<br/>"); } }} // Driver code let arr = [ "neveropen", "for", "neveropen" ]; let n = arr.length; insertNames(arr, n); // This code is contributed by susmitakundugoaldanga.</script> |
Output
No No Yes
Time Complexity: O(n), Where n is the length of the given string array.
Auxiliary Space: O(n), To store the names of the employees in a set.
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