Given a string S of size N consisting of unique ID and Domain Name of a unique website, the task is to find the ID and the Domain Name in the given string if the ID is of the form [char, char, char, char, char, digit, digit, digit, digit, char].
Examples:
Input: S = “We thank ABCDE1234F for visiting us and buying products item AMZrr@!k. For more offers, visit us at www.amazon.com”
Output:
ID = ABCDE1234F
Domain = amazon.comInput: S = “Hi PQRST5678D, it was a pleasure to host you. See www.oyo.com, our official website for future stays”
Output:
ID = PQRST5678D
Domain = oyo.com
Approach: The simplest approach to solve the given problem is to split the string into words and find if the split string is ID or Domain. Follow the steps below to solve the problem:
- First split the words of the string separated by space and store them in a vector of string say words[].
- Initialize two empty strings, say ID and Domain to store the resultant ID and Domain Name.
- Traverse the vector of string words[] and perform the following steps:
- Initialize a variable, say flag as false to store if the current string satisfies the format of ID or not.
- If the length of the current string is 10 and if the first 5 characters and the last character is non-alphabets or any of the remaining characters of the string is non-numeric then mark the flag as true.
- If the value of the flag is false, then assign the current string to ID.
- If the first substring of the current string, say SS over the range [0, 2] is “www” and substring over the range [SS.length() – 3, SS.length() – 1] is “com” then assign the domain name i.e., substring over the range [3, SS.length() – 1] to Domain.
- After completing the above steps, print the value of ID and Domain as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a character is// alphabet or notbool ischar(char x){ if ((x >= 'A' && x <= 'Z') || (x >= 'a' && x <= 'z')) { return 1; } return 0;}// Function to check if a character is// a numeric or notbool isnum(char x){ if (x >= '0' && x <= '9') return 1; return 0;}// Function to find ID and Domain// name from a given stringvoid findIdandDomain(string S, int N){ // Stores ID and the domain names string ID, Domain; // Stores the words of string S vector<string> words; // Stores the temporary word string curr = ""; // Traverse the string S for (int i = 0; i < N; i++) { // If the current character // is space if (S[i] == ' ') { // Push the curr in words words.push_back(curr); // Update the curr curr = ""; } // Otherwise else { if (S[i] == '.') { if (i + 1 == N || (i + 1 < N && S[i + 1] == ' ')) continue; } curr += S[i]; } } // If curr is not empty if (curr.length()) words.push_back(curr); for (string ss : words) { // If length of ss is 10 if (ss.size() == 10) { bool flag = 0; // Traverse the string ss for (int j = 0; j <= 9; j++) { // If j is in the range // [5, 9) if (j >= 5 && j < 9) { // If current character // is not numeric if (isnum(ss[j]) == 0) // Mark flag 1 flag = 1; } // Otherwise else { // If current character // is not alphabet if (ischar(ss[j]) == 0) // Mark flag 1 flag = 1; } } // If flag is false if (!flag) { // Assign ss to ID ID = ss; } } // If substring formed by the // first 3 character is "www" // and last 3 character is "moc" if (ss.substr(0, 3) == "www" && ss.substr( ss.length() - 3, 3) == "com") { // Update the domain name Domain = ss.substr( 4, ss.size() - 4); } } // Print ID and Domain cout << "ID = " << ID << endl; cout << "Domain = " << Domain;}// Driver Codeint main(){ string S = "We thank ABCDE1234F for visiting " "us and buying " "products item AMZrr@!k. For more " "offers, visit " "us at www.amazon.com"; int N = S.length(); findIdandDomain(S, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to check if a character is// alphabet or notstatic boolean ischar(char x){ if ((x >= 'A' && x <= 'Z') || (x >= 'a' && x <= 'z')) { return true; } return false;}// Function to check if a character is// a numeric or notstatic boolean isnum(char x){ if (x >= '0' && x <= '9') return true; return false;}// Function to find ID and Domain// name from a given Stringstatic void findIdandDomain(String S, int N){ // Stores ID and the domain names String ID = "", Domain = ""; // Stores the words of String S Vector<String> words = new Vector<String>(); // Stores the temporary word String curr = ""; // Traverse the String S for(int i = 0; i < N; i++) { // If the current character // is space if (S.charAt(i) == ' ') { // Push the curr in words words.add(curr); // Update the curr curr = ""; } // Otherwise else { if (S.charAt(i) == '.') { if (i + 1 == N || (i + 1 < N && S.charAt(i + 1) == ' ')) continue; } curr += S.charAt(i); } } // If curr is not empty if (curr.length() > 0) words.add(curr); for(String ss : words) { // If length of ss is 10 if (ss.length() == 10) { boolean flag = false; // Traverse the String ss for(int j = 0; j <= 9; j++) { // If j is in the range // [5, 9) if (j >= 5 && j < 9) { // If current character // is not numeric if (isnum(ss.charAt(j)) == false) // Mark flag 1 flag = true; } // Otherwise else { // If current character // is not alphabet if (ischar(ss.charAt(j)) == false) // Mark flag 1 flag = true; } } // If flag is false if (!flag) { // Assign ss to ID ID = ss; } } // If subString formed by the // first 3 character is "www" // and last 3 character is "moc" if (ss.length() > 2 && ss.substring(0, 3).equals("www") && ss.substring(ss.length() - 3).equals("com")) { // Update the domain name Domain = ss.substring(4, ss.length()); } } // Print ID and Domain System.out.print("ID = " + ID + "\n"); System.out.print("Domain = " + Domain);}// Driver Codepublic static void main(String[] args){ String S = "We thank ABCDE1234F for visiting " + "us and buying products item AMZrr@!k. " + "For more offers, visit us at www.amazon.com"; int N = S.length(); findIdandDomain(S, N);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to check if a character is# alphabet or notdef ischar(x): if ((x >= 'A' and x <= 'Z') or (x >= 'a' and x <= 'z')): return 1 return 0# Function to check if a character is# a numeric or notdef isnum(x): if (x >= '0' and x <= '9'): return 1 return 0# Function to find ID and Domain# name from a givendef findIdandDomain(S, N): # Stores ID and the domain names ID, Domain = "", "" # Stores the words of S words = [] # Stores the temporary word curr = "" # Traverse the S for i in range(N): # If the current character # is space if (S[i] == ' '): # Push the curr in words words.append(curr) # Update the curr curr = "" # Otherwise else: if (S[i] == '.'): if (i + 1 == N or (i + 1 < N and S[i + 1] == ' ')): continue curr += S[i] # If curr is not empty if (len(curr)): words.append(curr) for ss in words: # If length of ss is 10 if (len(ss) == 10): flag = 0 # Traverse the ss for j in range(10): # If j is in the range # [5, 9) if (j >= 5 and j < 9): # If current character # is not numeric if (isnum(ss[j]) == 0): # Mark flag 1 flag = 1 # Otherwise else: # If current character # is not alphabet if (ischar(ss[j]) == 0): # Mark flag 1 flag = 1 # If flag is false if (not flag): # Assign ss to ID ID = ss # If sub formed by the # first 3 character is "www" # and last 3 character is "moc" if (ss[0: 3] == "www" and ss[len(ss) - 3: ]== "com"): # Update the domain name Domain = ss[4: len(ss) ] # Print ID and Domain print("ID =", ID) print("Domain =", Domain)# Driver Codeif __name__ == '__main__': S = "We thank ABCDE1234F for visiting us "\ "and buying products item AMZrr@!k. "\ "For more offers, visit us at www.amazon.com" N = len(S) findIdandDomain(S, N)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG{// Function to check if a character is// alphabet or notstatic bool ischar(char x){ if ((x >= 'A' && x <= 'Z') || (x >= 'a' && x <= 'z')) { return true; } return false;}// Function to check if a character is// a numeric or notstatic bool isnum(char x){ if (x >= '0' && x <= '9') return true; return false;}// Function to find ID and Domain// name from a given Stringstatic void findIdandDoMain(String S, int N){ // Stores ID and the domain names String ID = "", Domain = ""; // Stores the words of String S List<String> words = new List<String>(); // Stores the temporary word String curr = ""; // Traverse the String S for(int i = 0; i < N; i++) { // If the current character // is space if (S[i] == ' ') { // Push the curr in words words.Add(curr); // Update the curr curr = ""; } // Otherwise else { if (S[i] == '.') { if (i + 1 == N || (i + 1 < N && S[i + 1] == ' ')) continue; } curr += S[i]; } } // If curr is not empty if (curr.Length > 0) words.Add(curr); foreach(String ss in words) { // If length of ss is 10 if (ss.Length == 10) { bool flag = false; // Traverse the String ss for(int j = 0; j <= 9; j++) { // If j is in the range // [5, 9) if (j >= 5 && j < 9) { // If current character // is not numeric if (isnum(ss[j]) == false) // Mark flag 1 flag = true; } // Otherwise else { // If current character // is not alphabet if (ischar(ss[j]) == false) // Mark flag 1 flag = true; } } // If flag is false if (!flag) { // Assign ss to ID ID = ss; } } // If subString formed by the // first 3 character is "www" // and last 3 character is "moc" if (ss.Length > 2 && ss.Substring(0, 3).Equals("www") && ss.Substring(ss.Length - 3).Equals("com")) { // Update the domain name Domain = ss.Substring(4, ss.Length-4); } } // Print ID and Domain Console.Write("ID = " + ID + "\n"); Console.Write("Domain = " + Domain);}// Driver Codepublic static void Main(String[] args){ String S = "We thank ABCDE1234F for visiting " + "us and buying products item AMZrr@!k. " + "For more offers, visit us at www.amazon.com"; int N = S.Length; findIdandDoMain(S, N);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the above approach// Function to check if a character is// alphabet or notfunction ischar(x){ if ((x >= 'A' && x <= 'Z') || (x >= 'a' && x <= 'z')) { return true; } return false;}// Function to check if a character is// a numeric or notfunction isnum(x){ if (x >= '0' && x <= '9') return true; return false;}// Function to find ID and Domain// name from a given stringfunction findIdandDomain(S, N){ // Stores ID and the domain names let ID, Domain; // Stores the words of string S let words = []; // Stores the temporary word let curr = ""; // Traverse the string S for(let i = 0; i < N; i++) { // If the current character // is space if (S[i] == ' ') { // Push the curr in words words.push(curr); // Update the curr curr = ""; } // Otherwise else { if (S[i] == '.') { if (i + 1 == N || (i + 1 < N && S[i + 1] == ' ')) continue; } curr += S[i]; } } // If curr is not empty if (curr.length >= 1) words.push(curr); for(let i = 0; i < words.length; i++) { // If length of ss is 10 if (words[i].length == 10) { let flag = 0; // Traverse the string ss for(let j = 0; j <= 9; j++) { // If j is in the range // [5, 9) if (j >= 5 && j < 9) { // If current character // is not numeric if (isnum(words[i][j]) == 0) // Mark flag 1 flag = 1; } // Otherwise else { // If current character // is not alphabet if (ischar(words[i][j]) == 0) // Mark flag 1 flag = 1; } } // If flag is false if (!flag) { // Assign ss to ID ID = words[i]; } } // If substring formed by the // first 3 character is "www" // and last 3 character is "moc" if (words[i].substring(0, 3) == "www" && words[i].substring( words[i].length - 3) == "com") { // Update the domain name Domain = words[i].substring(4); } } // Print ID and Domain document.write("ID = " + ID + "<br>"); document.write("Domain = " + Domain);}// Driver Codelet S = "We thank ABCDE1234F for visiting " + "us and buying products item AMZrr@!k. " + "For more offers, visit us at www.amazon.com";let N = S.length;findIdandDomain(S, N);// This code is contributed by Dharanendra L V.</script> |
Output:
ID = ABCDE1234F Domain = amazon.com
Time Complexity: O(N)
Auxiliary Space: O(N)
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!
