From caba50e4ed85e02acc7f90403b6d6f16a8861529 Mon Sep 17 00:00:00 2001
From: NageshOdedra <nageshh.odedra@gmail.com>
Date: Wed, 7 Aug 2024 09:39:36 +0530
Subject: [PATCH] added two new files

---
 check_anagrams.py  | 50 ++++++++++++++++++++++++++++++++++++++++++++++
 check_semiprime.py | 34 +++++++++++++++++++++++++++++++
 2 files changed, 84 insertions(+)
 create mode 100644 check_anagrams.py
 create mode 100644 check_semiprime.py

diff --git a/check_anagrams.py b/check_anagrams.py
new file mode 100644
index 0000000..2b3b857
--- /dev/null
+++ b/check_anagrams.py
@@ -0,0 +1,50 @@
+# Problem: Two strings of sizes m and n are given,
+# we have to find how many characters need to be
+# 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:
+        if CHARACTER_HASH[char] > 0:
+            common_letters += 1
+    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"
+
+    mapLettersToHash(text_1)
+    common = computeCommonLetters(text_2)
+    result = computeUncommonLetters(text_1, text_2, common)
+    print(result)
diff --git a/check_semiprime.py b/check_semiprime.py
new file mode 100644
index 0000000..3b48682
--- /dev/null
+++ b/check_semiprime.py
@@ -0,0 +1,34 @@
+# Context: A semiprime is a product of two prime
+# numbers, not necessarily distinct.
+# Squares of prime numbers are also semiprimes.
+
+# Problem: Find the numbers which are semiprimes,
+# within a given range. For e.g. 1 to 100.
+
+
+def isSemiprime(num):
+    # start with the smallest prime
+    prime = 2
+    # initialize counter to 0
+    count = 0
+    # Design of while loop:
+    # 1. if count exceeds 2, it is not a semiprime, e.g. 30 = 2*3*5
+    # 2. when the number becomes 1, we have found the second prime
+    while count < 3 and num != 1:
+        # if the number is divisible by current prime,
+        # increment count, else move to new prime
+        if not (num % prime):
+            num = num / prime
+            count = count + 1
+        else:
+            prime = prime + 1
+    # if count is two, given number is a semiprime
+    return count == 2
+
+
+for i in range(1, 100):
+    if isSemiprime(i):
+        print(i, end=" ")
+
+# Result: 4 6 9 10 14 15 21 22 25 26 33 34 35 38 39 46 49
+# 51 55 57 58 62 65 69 74 77 82 85 86 87 91 93 94 95