Validating and Parsing Email Addresses Hacker Rank Solution – Queslers

Problem : Validating and Parsing Email Addresses Hacker Rank Solution

A valid email address meets the following criteria:

• It’s composed of a username, domain name, and extension assembled in this format: username@domain.extension
• The username starts with an English alphabetical character, and any subsequent characters consist of one or more of the following: alphanumeric characters, -,., and _.
• The domain and extension contain only English alphabetical characters.
• The extension is 1, 2, or 3 characters in length.

Given n pairs of names and email addresses as input, print each name and email address pair having a valid email address on a new line.
Hint : Try using Email.utils() to complete this challenge. For example, this code:

import email.utils
print email.utils.parseaddr('DOSHI <DOSHI@hackerrank.com>')
print email.utils.formataddr(('DOSHI', 'DOSHI@hackerrank.com'))

produces this output:

('DOSHI', 'DOSHI@hackerrank.com')
DOSHI <DOSHI@hackerrank.com>

---

Input Format :

The first line contains a single integer, n, denoting the number of email address.
Each line i of the n subsequent lines contains a name and an email address as two space-separated values following this format:

name <user@email.com>

Constraints :

• 0 < N < 100

Output Format :

Print the space-separated name and email address pairs containing valid email addresses only. Each pair must be printed on a new line in the following format:

name <user@email.com>

You must print each valid email address in the same order as it was received as input.

---

Sample Input :

2  
DEXTER <dexter@hotmail.com>
VIRUS <virus!@variable.:p>

Sample Output :

DEXTER <dexter@hotmail.com>

Explanation :

dexter@hotmail.com is a valid email address, so we print the name and email address pair received as input on a new line.
virus!@variable.:p is not a valid email address because the username contains an exclamation point (!) and the extension contains a colon (:). As this email is not valid, we print nothing.

Validating and Parsing Email Addresses Hacker Rank Solution in python 2

# Enter your code here. Read input from STDIN. Print output to STDOUT
import re
N = input()
for i in xrange(N): 
    s = raw_input()
    a = re.search(r'<[a-zA-Z][\w\.-]*@[a-zA-Z]*\.[a-zA-Z]{1,3}>', s)
    if bool(a):
        print(s)

Validating and Parsing Email Addresses Hacker Rank Solution in python 3

# Enter your code here. Read input from STDIN. Print output to STDOUT
import re
n = int(input())
for _ in range(n):
    x, y = input().split(' ')
    m = re.match(r'<[A-Za-z](\w|-|\.|_)+@[A-Za-z]+\.[A-Za-z]{1,3}>', y)
    if m:
        print(x,y)

Validating and Parsing Email Addresses Hacker Rank Solution in pypy

import email.utils
import re
n = int(raw_input())
for i in range(n):
    p = email.utils.parseaddr(raw_input())
    if re.match(r'[a-z][\w.-]+@[a-z]+\.[a-z]{1,3}$', p[1], re.I):
        print email.utils.formataddr(p)

Validating and Parsing Email Addresses Hacker Rank Solution in pypy 3

# Enter your code here. Read input from STDIN. Print output to STDOUT

import re
import email.utils
n = int(input().strip())
temp =[]
a = re.compile(r'<[a-z0-9][\w._-]+@[a-z]+\.[a-z]{1,3}>', re.I)
for _ in range(n):
    temp.append(input().strip())
for i, x in enumerate(temp):
    v =  a.search(x)
    if v:
        print(temp[i])

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