Tuesday, December 3, 2019

Regex Substitution - Hacker Rank Solution

The re.sub() tool (sub stands for substitution) evaluates a pattern and, for each valid match, it calls a method (or lambda).
The method is called for all matches and can be used to modify strings in different ways.

The re.sub() method returns the modified string as an output.
Learn more about .
Transformation of Strings
Code
import re

#Squaring numbers
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

Replacements in Strings
Code
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) #remove comment
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>

Task
You are given a text of  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.
Input Format
The first line contains the integer, .
The next  lines each contain a line of the text.
Constraints
Neither && nor || occur in the start or end of each line.
Output Format
Output the modified text.
Sample Input
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.
Sample 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.

Regex Substitution - Hacker Rank Solution

One can solve this challenge by using the re.sub() tool along with (?<= ) which is a positive lookbehind that makes sure that the pattern is preceded by a space. Also, by using (?= ) which is a positive lookahead that makes sure that the pattern is followed by a space.

import re

for line in range(int(raw_input())):
    string = ''
    string = re.sub(r'(?<= )&&(?= )','and',raw_input())
    string = re.sub(r'(?<= )\|\|(?= )','or',string)
    print string

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