Update check_anagrams.py

Author: Amogh Singhal

check_anagrams.py

@@ -3,18 +3,29 @@
 # removed from both the string so that they become
 # anagrams of each other
 
+# Anagrams: Words that are made from rearranging the letters of another word
+
+# Algorithm: We will use dictionaries to keep track of characters. 
+# The idea is to get the common occuring characters and derive 
+# uncommon characters
+
 import string
 
+# letters will be a string of the form "abc...xyz"
+# CHARACTER_HASH looks like this {'a'0, 'b':0 ...., 'z':0}
 letters = string.ascii_lowercase
 CHARACTER_HASH = dict(zip(letters, [0] * len(letters)))
 
 
+# This method will mark all the letters occuring in 'text_a'
 def mapLettersToHash(text_a):
     for char in text_a:
         if char in CHARACTER_HASH.keys():
             CHARACTER_HASH[char] += 1
 
 
+# This method will count the letters present in 'text_b', also found in 'text_a'
+# These will be charcaters whose frequency in HASH is greater than zero
 def computeCommonLetters(text_b):
     common_letters = 0
     for char in text_b:
@@ -23,14 +34,17 @@ def computeCommonLetters(text_b):
     return common_letters
 
 
+# Now we derive how many uncommon letters are present,
+# This is done by subtracting twice the count of common letters
+# from the total length of both the strings
 def computeUncommonLetters(text_a, text_b, common_letters):
     return abs(len(text_a) + len(text_b) - (2 * common_letters))
 
+if __name__ == "__main__":
+    text_1 = "hello"
+    text_2 = "billion"
 
-text_1 = "hello"
-text_2 = "billion"
-
-mapLettersToHash(text_1)
-common = computeCommonLetters(text_2)
-result = computeUncommonLetters(text_1, text_2, common)
-print(result)
+    mapLettersToHash(text_1)
+    common = computeCommonLetters(text_2)
+    result = computeUncommonLetters(text_1, text_2, common)
+    print(result)