Python13_Regex_and_Parsing

## Python
## Regex and Parsing


## Detect Floating Point Number
"""
eg, 
4.0O0 = False (there is a letter 'O', not zero)
-1.00 = True
+4.54 = True
SomeRandomStuff = False
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT
# Condensed version:
import re
for _ in range(int(input())):
    # print(bool(re.match(r'^[-+]?[0-9]*\.[0-9]+$', input())))
    # print(bool(re.search(r'^[+-]?\d{0,}\.\d{1,}$',input())))
    print(bool(re.match(r'^[-+]?\d*\.\d+$', input())))

# # Simple readable version:
# import re
# n = int(input())
# for _ in range(n):
#     str = input()
#     if (re.match(r'^[-+]?[0-9]*\.[0-9]+$', str)):
#         print(True)
#     else:
#         print(False)


## Re.split()
"""
Spliting 100,000,000.000 to 
100
000
000
000
"""
regex_pattern = r'[.,]+'	# Do not delete 'r'.
import re
print("\n".join(re.split(regex_pattern, input())))

# # Option 2
# import re
# print(*filter(None, re.split(r'[.,]+', input())), sep='\n')

# # Option 3
# import re
# print(*re.split(r'[.,]+', input().strip('., ')), sep='\n')

# # Option 4
# import re
# print(*re.split('[,.]+', input()), sep='\n')


## Group(), Groups() & Groupdict()
"""
>>> import re
>>> m = re.match(r'(\w+)@(\w+)\.(\w+)','username@hackerrank.com')
>>> m.group(0)       # The entire match 
'username@hackerrank.com'
>>> m.group(1)       # The first parenthesized subgroup.
'username'
>>> m.group(2)       # The second parenthesized subgroup.
'hackerrank'
>>> m.group(3)       # The third parenthesized subgroup.
'com'
>>> m.group(1,2,3)   # Multiple arguments give us a tuple.
('username', 'hackerrank', 'com')
>>> m.groups()
('username', 'hackerrank', 'com')

>>> m = re.match(r'(?P<user>\w+)@(?P<website>\w+)\.(?P<extension>\w+)','myname@hackerrank.com')
>>> m.groupdict()
{'website': 'hackerrank', 'user': 'myname', 'extension': 'com'}
"""
# Given string S, find the first occurrence of an alphanumeric character in S.
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# import re
# m = re.search(r'([a-zA-Z0-9])\1+', input().strip())
# print(m.group(1) if m else -1)

# # Option 2
# s=list(input())
# for i in range(len(s)):
#     try:
#         if s[i]==s[i+1]:
#             if  s[i].isalnum():
#                 print(s[i])
#                 break
#     except:
#         print("-1")

# Option 3
import re
reg = re.match(r".*?([a-zA-Z0-9]+)\1", input())
if reg:
    print(reg.group(1))
else:
    print(-1)


## Re.findall() & Re.finditer()
"""
The expression re.findall() returns all the non-overlapping matches of patterns in a string as a list of strings.
>>> import re
>>> re.findall(r'\w','http://www.hackerrank.com/')
['h', 't', 't', 'p', 'w', 'w', 'w', 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k', 'c', 'o', 'm']

The expression re.finditer() returns an iterator yielding MatchObject instances over all non-overlapping matches for the re pattern in the string.
>>> import re
>>> re.finditer(r'\w','http://www.hackerrank.com/')
<callable-iterator object at 0x0266C790>
>>> map(lambda x: x.group(),re.finditer(r'\w','http://www.hackerrank.com/'))
['h', 't', 't', 'p', 'w', 'w', 'w', 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k', 'c', 'o', 'm']

given a string S consisting of alphanumeric characters, spaces and symbols(+,-).
Your task is to find all the substrings of S that contains 2 or more vowels, located in between 2 consonants
eg. S = 'rabcdeefgyYhFjkIoomnpOeorteeeeet'
Output:
ee
Ioo
Oeo
eeeee
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# import re
# v = 'aeiou'
# c = 'qwrtypsdfghjklzxcvbnm'
# m = re.findall(r'(?<=[%s])([%s]{2,})[%s]' % (c, v, c), input(), flags = re.I)
# print('\n'.join(m or ['-1']))

# # Option 2
# import re
# c = '[qwrtypsdfghjklzxcvbnm]'
# m = re.findall('(?<=' + c +')([aeiou]{2,})' + c, input(), re.I)
# print('\n'.join(m or ['-1']))

# # Option 3
# import re
# regex = re.findall('(?<=[^AEIOUaeiou])([AEIOUaeiou]{2,})[^AEIOUaeiou]', input().strip())
# print('\n'.join(regex) if regex else -1)

# Option 4
import re
x = input() 
v = re.findall(r'(?<=[^AEIOUaeiou])([AEIOUaeiou]{2,})[^AEIOUaeiou]', x)
y = ('\n'.join(v))
if v: 
    print(y) 
else: 
    print(-1)
"""
v =    # assign to v, which is a list
re.findall(r'   # to find all relevant format
(?<=[^AEIOUaeiou])    # any char in front that is not vowel
([AEIOUaeiou]{2,})    # capture {2,} ie, 2 or more vowels
[^AEIOUaeiou]    # not vowel
', x)     # find all relevant format in input str x
"""


## Re.start() & Re.end()
"""
These expressions return the indices of the start and end of the substring matched by the group.
>>> import re
>>> m = re.search(r'\d+','1234')
>>> m.end()
4
>>> m.start()
0

You are given a string S.
Your task is to find the indices of the start and end of string k in S.
eg. 
aaadaa
aa
Output:
(0, 1)  
(1, 2)
(4, 5)
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# S = input()
# k = input()
# import re
# pattern = re.compile(k)
# r = pattern.search(S)
# if not r: 
#     print('(-1, -1)')
# while r:
#     print('({0}, {1})'.format(r.start(), r.end() - 1))
#     r = pattern.search(S, r.start() + 1)

# Option 2
import re 
S = input()
k = input()
if re.search(k, S)==None:    # none matches
    print( (-1, -1) )
else:      # >=1 matches
    for i in range(len(S)-len(k)+1):
        if S[i:i+len(k)]==k:
            print((i, i+len(k)-1))


## Regex Substitution
"""
The re.sub() method returns the modified string as an output.
re.sub( a, becomes b, input)
#Squaring numbers
import re
def square(match):
    number = int(match.group(0))
    return str(number**2)
print re.sub(r"\d+", square, "1 2 3 4 5 6 7 8 9")

Output:
1 4 9 16 25 36 49 64 81

#remove comment
import re
html = """
<head>
<title>HTML</title>
</head>
<object type="application/x-flash" 
  data="your-file.swf" 
  width="0" height="0">
  <!-- <param name="movie"  value="your-file.swf" /> -->
  <param name="quality" value="high"/>
</object>  """
print re.sub("(<!--.*?-->)", "", html) 

Output
<head>
<title>HTML</title>
</head>
<object type="application/x-flash" 
  data="your-file.swf" 
  width="0" height="0">

  <param name="quality" value="high"/>
</object>

You are given a text of N lines. The text contains && and || symbols.
Your task is to modify those symbols to the following:
&& →  and
|| →  or
Both && and || should have a space " " on both sides

Example:
11
a = 1;
b = input();

if a + b > 0 && a - b < 0:
    start()
elif a*b > 10 || a/b < 1:
    stop()
print set(list(a)) | set(list(b)) 
#Note do not change &&& or ||| or & or |
#Only change those '&&' which have space on both sides.
#Only change those '|| which have space on both sides (this has space on 1 side).

Output:
a = 1;
b = input();

if a + b > 0 and a - b < 0:
    start()
elif a*b > 10 or a/b < 1:
    stop()
print set(list(a)) | set(list(b)) 
#Note do not change &&& or ||| or & or |
#Only change those '&&' which have space on both sides.
#Only change those '|| which have space on both sides.  

"""
# # Option 1
# import re
# N = int(input())
# for i in range(N):
#     print(re.sub(r'(?<= )(&&|\|\|)(?= )', lambda x: 'and' if x.group() == '&&' else 'or', input()))
# # (&&|\|\|) means (&& or ||)
# # since | is a special character in re, escape each of them \|\| to get ||

# # Option 2
# import re
# for _ in range(int(input())):
#     s = input()
#     s = re.sub(r' &&(?= )', ' and', s)
#     s = re.sub(r' \|\|(?= )', ' or', s)
#     print(s)

# Option 3
import re
def f(x): 
    return 'and' if x.group(0) == '&&' else 'or'
for _ in range(int(input())):
    s = input()
    print(re.sub(r'(?<= )([&|])\1(?= )', f, s))


## Validating Roman Numerals
# You are given a string, and you have to validate whether it's a valid Roman numeral. If it is valid, print True. Otherwise, print False. Roman numeral between 1 and 3999.

# # Option 1
# regex_pattern = r"(?<=^)M{0,3}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})(?=$)"

# # Option 2
# regex_pattern = r"(M{1,3})?((C{1,3})|C(D|M)|D(C{0,3})|M)?((X{1,3})|X(C|L)|L(X{0,3})|C)?((I{1,3})|I(V|X)|V(I{0,3})|X)\b"

# # Option 3
# regex_pattern = r"^(M{0,3}CMXCIX|M{0,3})(D{0,1}?C{0,3}|CD)(L{0,1}?X{0,3}|XL)(V{0,1}?I{0,3}|IV)$"

# Option 4
# thousand = M{0,3}
# hundred = (C[MD]|D?C{0,3})
# ten = (X[CL]|L?X{0,3})
# digit = (I[VX]|V?I{0,3})
regex_pattern = r"^M{0,3}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$"

import re
print(str(bool(re.match(regex_pattern, input()))))


## Validating phone numbers
# A valid mobile number is a ten digit number starting with a 7, 8 or 9.

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

# # Option 1
# # [789]\d{9}$ checks 1st digit is 7,8,9, plus next 9 are digits (total 10 digits)
# import re
# [print('YES' if re.match(r'[789]\d{9}$',input()) else 'NO') for _ in range(int(input()))]

# # Option 2
# import re
# for i in range(int(input())):
#     if re.match(r'[789]\d{9}$', input()):   
#         print('YES') 
#     else:  
#         print('NO')

# # Option 3
# import re
# for _ in range(int(input())):
#     s = input().strip()
#     print('YES' if (bool(re.search(r'^[789]', s)) and bool(re.search(r'^(\d){10}$',s))) else 'NO')

# Option 4 (without regex)
check_list = '789'
for i in range(int(input())):
    num = input()
    if num.isdigit() and num[0] in check_list and len(num)==10:
        print('YES') 
    else:
        print('NO')


## Validating and Parsing Email Addresses
"""
A valid email address is composed of a username, domain name, and extension assembled in this format: username@domain.extension
example:
import email.utils
print email.utils.parseaddr('DOSHI <DOSHI@hackerrank.com>')
print email.utils.formataddr(('DOSHI', 'DOSHI@hackerrank.com'))

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

Input n lines of name and an email address as two space-separated values following this format:
name <user@email.com>
Output to 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>

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

Sample Output:
DEXTER <dexter@hotmail.com>
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# Note: alphanumeric \w includes [A-Za-z0-9_]

# # Option 1
# 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)

# Option 2
import re
import email.utils as email
n = int(input())
regex = r'^[a-zA-Z][\w\.-]+@[a-zA-Z]+\.[a-zA-Z]{1,3}$'
for _ in range(n):
    e = input()
    p = email.parseaddr(e)
    if re.search(regex, p[1]):
        print(email.formataddr(p))


## Hex Color Code
"""
You are given N lines of CSS code. Your task is to print all valid Hex Color Codes, in order of their occurrence from top to bottom.

Sample Input:
11
#BED
{
    color: #FfFdF8; background-color:#aef;
    font-size: 123px;
    background: -webkit-linear-gradient(top, #f9f9f9, #fff);
}
#Cab
{
    background-color: #ABC;
    border: 2px dashed #fff;
}   

Sample Output:
#FfFdF8
#aef
#f9f9f9
#fff
#ABC
#fff
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# Option 1
import re
N = int(input())
for i in range(N):
    s = input()
    x = s.split()
    if len(x)>1  and  '{' not in x:
        x = re.findall(r'#[a-fA-F0-9]{3,6}',s)
        [print(i) for i in x]

# # Option 2
# import re, sys
# [print(j) for i in sys.stdin for j in re.findall('[\s:](#[a-f0-9]{6}|#[a-f0-9]{3})', i, re.I)]


## HTML Parser - Part 1
"""
An HTMLParser instance is fed HTML data and calls handler methods when start tags, end tags, text, comments, and other markup elements are encountered.
You are given an HTML code snippet of  lines.
Your task is to print start tags, end tags and empty tags separately.

Sample Input:
2
<html><head><title>HTML Parser - I</title></head>
<body data-modal-target class='1'><h1>HackerRank</h1><br /></body></html>

Sample Output:
Start : html
Start : head
Start : title
End   : title
End   : head
Start : body
-> data-modal-target > None
-> class > 1
Start : h1
End   : h1
Empty : br
End   : body
End   : html
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# from html.parser import HTMLParser
# class MyHTMLParser(HTMLParser):
#     def handle_starttag(self, tag, attrs):        
#         print('Start :', tag)
#         for ele in attrs:
#             print('->',ele[0],'>',ele[1])
#     def handle_endtag(self, tag):
#         print('End   :', tag)
#     def handle_startendtag(self, tag, attrs):
#         print('Empty :', tag)
#         for ele in attrs:
#             print('->',ele[0],'>',ele[1])
# parser = MyHTMLParser()
# for _ in range(int(input())):
#     parser.feed(input())

# Option 2
from html.parser import HTMLParser
class MyHTMLParser(HTMLParser):
    def handle_starttag(self, tag, attrs):
        print('Start :', tag)
        self.value(attrs)
    def handle_endtag(self, tag):
        print('End   :', tag)
    def handle_startendtag(self, tag, attrs):
        print('Empty :', tag)
        self.value(attrs)
    def value(self, attrs = None):
        if attrs:
            [print('->', attr, '>', val) for attr, val, in attrs]
parser = MyHTMLParser()
for _ in range(int(input())):
    parser.feed(input())


## HTML Parser - Part 2
"""
You are given an HTML code snippet of N lines.
Your task is to print the single-line comments, multi-line comments and the data.
Print the result in the following format:
>>> Single-line Comment  
Comment
>>> Data                 
My Data
>>> Multi-line Comment  
Comment_multiline[0]
Comment_multiline[1]
>>> Data
My Data
>>> Single-line Comment:  

Sample Input:
4
<!--[if IE 9]>IE9-specific content
<![endif]-->
<div> Welcome to HackerRank</div>
<!--[if IE 9]>IE9-specific content<![endif]-->

Sample Output:
>>> Multi-line Comment
[if IE 9]>IE9-specific content
<![endif]
>>> Data
 Welcome to HackerRank
>>> Single-line Comment
[if IE 9]>IE9-specific content<![endif]
"""
# # Option 1
# from html.parser import HTMLParser
# class MyHTMLParser(HTMLParser):
#     def handle_comment(self, comment):
#         number_of_line = len(comment.split('\n'))
#         if number_of_line > 1:
#             print('>>> Multi-line Comment')
#         else:
#             print('>>> Single-line Comment')
#         print(comment)
#     def handle_data(self, data):
#         if data.strip():
#             print(">>> Data")
#             print(data)
# parser = MyHTMLParser()
# html_string = ''
# for i in range(int(input())):
#     html_string += input().rstrip() + '\n'
# parser.feed(html_string)
# parser.close()

# Option 2
from html.parser import HTMLParser
class MyHTMLParser(HTMLParser):
    def handle_comment(self, comment):
        if '\n' in comment:
            print('>>> Multi-line Comment')
        else:
            print('>>> Single-line Comment')
        print(comment)
    def handle_data(self, data):
        if len(data) > 1: 
            print('>>> Data', data, sep='\n')
parser = MyHTMLParser()
html_string = ''
for i in range(int(input())):
    html_string += input().rstrip() + '\n'
parser.feed(html_string)
parser.close()


## Detect HTML Tags, Attributes and Attribute Values
"""
The first line contains an integer , the number of lines in the HTML code snippet.
The next  lines contain HTML code.
Print the HTML tags, attributes and attribute values in order of their occurrence from top to bottom in the snippet.
Format your answers as explained in the problem statement.

Sample Input:
9
<head>
<title>HTML</title>
</head>
<object type="application/x-flash" 
  data="your-file.swf" 
  width="0" height="0">
  <!-- <param name="movie" value="your-file.swf" /> -->
  <param name="quality" value="high"/>
</object>

Sample Output:
head
title
object
-> type > application/x-flash
-> data > your-file.swf
-> width > 0
-> height > 0
param
-> name > quality
-> value > high
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# from html.parser import HTMLParser
# class MyHTMLParser(HTMLParser):
#     def handle_starttag(self, tag, attrs):
#         print(tag)
#         if len(attrs) > 0:
#             for e in attrs:
#                 print(f"-> {e[0]} > {e[1]}")
# html = ''
# for line in range(int(input())):
#     html += input().rstrip()
#     html += '\n'
# parser = MyHTMLParser()
# parser.feed(html)

# Option 2
from html.parser import HTMLParser
class MyHTMLParser(HTMLParser):
    def handle_starttag(self, tag, attrs):        
        print(tag)
        for e in attrs:
            print ('->', e[0], '>', e[1])
parser = MyHTMLParser()
for _ in range(int(input())):
    parser.feed(input()) 

# # Option 3
# from html.parser import HTMLParser
# class MyHTMLParser(HTMLParser):
#     def handle_starttag(self, tag, attrs):
#       print(tag)
#       for e in attrs:
#         print("-> {} > {}".format(e[0],e[1]))
# MyParser = MyHTMLParser()
# for i in range(int(input())):
#     MyParser.feed(input())


## Validating UID (unique identification number)
"""
The company has assigned you the task of validating all the randomly generated UIDs.
A valid UID must follow the rules below:
It must contain at least 2 uppercase English alphabet characters.
It must contain at least 3 digits (0 - 9).
It should only contain alphanumeric characters (a - z, A - Z & 0 - 9).
No character should repeat.
There must be exactly 10 characters in a valid UID.

Sample Input:
2
B1CD102354
B1CDEF2354

Sample Output:
Invalid
Valid
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# import re
# p = re.compile(r'[^a-zA-Z0-9]|([a-zA-Z0-9]).*?(?=\1)|^[a-zA-Z0-9]{,9}$|' + \
#                 '^[a-zA-Z0-9]{11,}|(?:^[^A-Z]*?[A-Z]?[^A-Z]*$)|' + \
#                 '(?:^[^0-9]*?[0-9]?[^0-9]*?[0-9]?[^0-9]*$)')
# n = int(input())
# for _ in range(n):
#     in_ = input()
#     print('Invalid' if p.search(in_) else 'Valid')

# Option 2
def check(message):
    if sum(1 for c in message if c.isupper()) <2:
        return 'Invalid'
    elif sum(1 for c in message if c.isdigit()) <3:
        return 'Invalid'
    elif sum(1 for c in message if not c.isalnum()) >0:
        return 'Invalid'
    elif len(set(message)) != 10:   # no repeat character
        return 'Invalid'
    else:
        return 'Valid'
n = int(input()) 
for i in range(n): 
    s = input() 
    print(check(s))


## Validating Credit Card Numbers
"""
A valid credit card from ABCD Bank has the following characteristics:
► It must start with a 4, 5 or 6.
► It must contain exactly 16 digits.
► It must only consist of digits (0-9).
► It may have digits in groups of 4, separated by one hyphen "-".
► It must NOT use any other separator like ' ' , '_', etc.
► It must NOT have 4 or more consecutive repeated digits.
Examples:
Valid Credit Card Numbers
4253625879615786
4424424424442444
5122-2368-7954-3214
Invalid Credit Card Numbers
42536258796157867       #17 digits in card number → Invalid 
4424444424442444        #Consecutive digits are repeating 4 or more times → Invalid
5122-2368-7954 - 3214   #Separators other than '-' are used → Invalid
44244x4424442444        #Contains non digit characters → Invalid
0525362587961578        #Doesn't start with 4, 5 or 6 → Invalid

Sample Input
6
4123456789123456
5123-4567-8912-3456
61234-567-8912-3456
4123356789123456
5133-3367-8912-3456
5123 - 3567 - 8912 - 3456

Sample Output
Valid
Valid
Invalid
Valid
Invalid
Invalid
"""
# Enter your code here. Read input from STDIN. Print output to STDOUT

# # Option 1
# import re
# regex = re.compile(
#     r'^'
#     r'(?!.*(\d)(-?\1){3})'
#     r'[456]'
#     r'\d{3}'
#     r'(?:-?\d{4}){3}'
#     r'$')
# for _ in range(int(input().strip())):
#     print('Valid' if regex.search(input().strip()) else 'Invalid')

# Option 2
import re
regex = re.compile(
    r'^'
    r'(?!.*(\d)(-?\1){3})'
    r'[456]'
    r'\d{3}'
    r'(?:-?\d{4}){3}'
    r'$')
for _ in range(int(input())):
    x = input()
    if regex.search(x):
        print('Valid')
    else:
        print('Invalid')


## Validating Postal Codes
"""
A valid postal code P have to fullfil both below requirements:
P must be a number in the range from 100000 to 999999 inclusive.
P must not contain more than one alternating repetitive digit pair.
Alternating repetitive digits are digits which repeat immediately after the next digit. In other words, an alternating repetitive digit pair is formed by two equal digits that have just a single digit between them.
For example:
121426 # Here, 1 is an alternating repetitive digit.
523563 # Here, NO digit is an alternating repetitive digit.
552523 # Here, both 2 and 5 are alternating repetitive digits.
"""
# Option 1
regex_integer_in_range = r'^[1-9][\d]{5}$'
regex_alternating_repetitive_digit_pair = r'(\d)(?=\d\1)'

# Option 2
regex_integer_in_range = r'^[1-9]\d{5}$'
regex_alternating_repetitive_digit_pair = r'(.)(?=.\1)'

# Option 3
regex_integer_in_range = r'\d{6}(?!\d)'
regex_alternating_repetitive_digit_pair = r'(.)(?=.\1)'

import re
P = input()

print (bool(re.match(regex_integer_in_range, P)) 
and len(re.findall(regex_alternating_repetitive_digit_pair, P)) < 2)


## Matrix Script
#!/bin/python3

import math
import os
import random
import re
import sys

first_multiple_input = input().rstrip().split()
n = int(first_multiple_input[0])
m = int(first_multiple_input[1])
matrix = []
for _ in range(n):
    matrix_item = input()
    matrix.append(matrix_item)

# # Option 1
# new = []
# for i in range(m):
#     new.append(''.join(list(map(lambda x:x[i],matrix))))
# print(re.sub(r'(?<=\w)[\s!@#$%&]+(?=\w)',' ',''.join(new)))

# # Option 2
# new = ''.join(list(map(lambda t : ''.join(t), zip(*matrix))))
# print(re.sub(r'(?<=[a-zA-Z0-9])[^a-zA-Z0-9]+(?=[a-zA-Z0-9])', ' ', new))

# Option 3
new = list()
for i in range(m):
    for j in range(n):
        new.append(matrix[j][i])
regex = re.compile('(?<=[a-z])[!@#\$%& ]+(?=[a-z])', flags = re.I)
print(regex.sub(' ', ''.join(new)))


## end ##