Given an array arr[] consisting of N strings where each string represents an email address consisting of English alphabets, ‘.’, ‘+’ and ‘@’, the task is to count the number of distinct emails present in the array according to the following rules:
- An email address can be split into two substrings, the prefix and suffix of ‘@’, which are the local name and domain name respectively.
- The ‘.’ character in the string in the local name is ignored.
- In the local name, every character after ‘+‘ is ignored.
Examples:
Input: arr[] = {“raghav.agg@neveropen.com”, “raghavagg@neveropen.com”}
Output: 1
Explanation: Removing all the ‘.’s before ‘@’ modifies the strings to {“raghavagg@neveropen.com”, “raghavagg@neveropen.com”}. Therefore, the total number of distinct emails present in the string are 1.Input: arr[] = {“avruty+dhir+gfg@neveropen.com”, “avruty+gfg@neveropen.com”, “av.ruty@neveropen.com”}
Output: 1
Approach: The given problem can be solved by storing each email in a HashSet after populating it according to the given rule and print the size of the HashSet obtained. Follow the steps below to solve the problem:
- Initialize a HashSet, say S, to store all the distinct strings after populating according to the given rules.
- Traverse the given array arr[] and perform the following steps:
- Find the position of ‘@’ and store it in a variable, say pos2.
- Delete all the ‘.’ characters before pos2 using erase() function.
- Update the position of ‘@’ i.e., pos2 = find(‘@’) and find the position of ‘+’ and store it in a variable say pos1 as S.find(‘+’).
- Now, erase all the characters after pos1 and before pos2.
- Insert all the updated strings in a HashSet S.
- After completing the above steps, print the size of HashSet S as the result.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count all the distinct // emails after preprocessing according // to the given rules int distinctEmails(vector<string>& emails) { // Traverse the given array of // strings arr[] for (auto& x : emails) { // Stores the position of '@' // in the string auto pos2 = x.find('@'); // If pos2 < x.size() if (pos2 < x.size()) // Erases all the occurrences // of '.' before pos2 x.erase( remove(x.begin(), x.begin() + pos2, '.'), x.begin() + pos2); // Stores the position of the // first '+' auto pos1 = x.find('+'); // Update the position pos2 pos2 = x.find('@'); // If '+' exists then erase // characters after '+' and // before '@' if (pos1 < x.size() and pos2 < x.size()) { x.erase(pos1, pos2 - pos1); } } // Insert all the updated strings // inside the set unordered_set<string> ans( emails.begin(), emails.end()); // Return the size of set ans return ans.size(); } // Driver Code int main() { vector<string> arr = { "raghav.agg@neveropen.com", "raghavagg@neveropen.com" }; // Function Call cout << distinctEmails(arr); return 0; } |
Java
/*package whatever //do not write package name here */// Java program for the above approachimport java.util.*;class GFG { // Function to count all the distinct // emails after preprocessing according // to the given rules static int distinctEmails(String emails[]) { HashSet<String> ans = new HashSet<>(); // Traverse the given array of // strings arr[] for (String x : emails) { // Stores the position of '@' // in the string int pos2 = x.indexOf('@'); // If pos2 < x.size() if (pos2 < x.length()) { // Erases all the occurrences // of '.' before pos2 String p = x.substring(0, pos2); p = p.replace(".", ""); x = p + x.substring(pos2); // Stores the position of the // first '+' int pos1 = x.indexOf('+'); // Update the position pos2 pos2 = x.indexOf('@'); // If '+' exists then erase // characters after '+' and // before '@' if (pos1 > 0 && pos1 < x.length() && pos2 < x.length()) x = x.substring(0, pos1) + x.substring(pos2); // Insert all the updated strings // inside the set ans.add(x); } } // Return the size of set ans return ans.size(); } // Driver Code public static void main(String args[]) { String arr[] = { "raghav.agg@neveropen.com", "raghavagg@neveropen.com" }; // Function Call System.out.println(distinctEmails(arr)); }}// contributed by akashish__ |
Python3
# Python3 program for the above approach# Function to count all the distinct# emails after preprocessing according# to the given rulesdef distinctEmails(emails): ans = set([]) # Traverse the given array of # strings arr[] for x in emails: # Stores the position of '@' # in the string pos2 = x.find('@') # If pos2 < x.size() if (pos2 < len(x)): # Erases all the occurrences # of '.' before pos2 p = x[:pos2] p = p.replace(".", "") x = p + x[pos2:] # Stores the position of the # first '+' pos1 = x.find('+') # Update the position pos2 pos2 = x.find('@') # If '+' exists then erase # characters after '+' and # before '@' if (pos1 > 0 and pos1 < len(x) and pos2 < len(x)): x = x[:pos1] + x[pos2:] # Insert all the updated strings # inside the set ans.add(x) # Return the size of set ans return len(ans)# Driver Codeif __name__ == "__main__": arr = ["raghav.agg@neveropen.com", "raghavagg@neveropen.com"] # Function Call print(distinctEmails(arr))# This code is contributed by ukasp |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG { // Function to count all the distinct // emails after preprocessing according // to the given rules static int distinctEmails(String[] emails) { HashSet<String> ans = new HashSet<String>(); // Traverse the given array of // strings arr[] for (int i = 0; i < emails.Length; i++) { string x = emails[i]; // Stores the position of '@' // in the string int pos2 = x.IndexOf('@'); // If pos2 < x.size() if (pos2 < x.Length) { // Erases all the occurrences // of '.' before pos2 String p = x.Substring(0, pos2); p = p.Replace(".", ""); x = p + x.Substring(pos2); // Stores the position of the // first '+' int pos1 = x.IndexOf('+'); // Update the position pos2 pos2 = x.IndexOf('@'); // If '+' exists then erase // characters after '+' and // before '@' if (pos1 > 0 && pos1 < x.Length && pos2 < x.Length) x = x.Substring(0, pos1) + x.Substring(pos2); // Insert all the updated strings // inside the set ans.Add(x); } } // Return the size of set ans return ans.Count; } // Driver Code static public void Main() { String[] arr = { "raghav.agg@neveropen.com", "raghavagg@neveropen.com" }; // Function Call Console.WriteLine(distinctEmails(arr)); }}// contributed by akashish__ |
Javascript
// Function to count all the distinct// emails after preprocessing according// to the given rulesfunction distinctEmails(emails) { var ans = new Set([]); // Traverse the given array of // strings arr[] for (var x of emails) { // Stores the position of '@' // in the string var pos2 = x.indexOf('@'); // If pos2 < x.size() if (pos2 < x.length) { // Erases all the occurrences // of '.' before pos2 let p = x.substring(0, pos2); p = p.replace(/\./g, ''); x = p + x.substring(pos2); // Stores the position of the // first '+' var pos1 = x.indexOf('+'); // Update the position pos2 pos2 = x.indexOf('@'); // If '+' exists then erase // characters after '+' and // before '@' if (pos1 > 0 && pos1 < x.length && pos2 < x.length) { x = x.substring(0, pos1) + x.substring(pos2); } // Insert all the updated strings // inside the set ans.add(x); } } // Return the size of set ans return ans.size;}// Driver Codeconst arr = [ 'raghav.agg@neveropen.com', 'raghavagg@neveropen.com',];// Function Callconsole.log(distinctEmails(arr));// This code is contributed by aadityaburujwale. |
Output:
1
Time Complexity: O(N2)
Auxiliary Space: O(N)
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