pwc241 solution in python

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diff --git a/challenge-241/pokgopun/python/ch-1.py b/challenge-241/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-241/
+"""
+
+Task 1: Arithmetic Triplets
+
+Submitted by: [42]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array (3 or more members) of integers in increasing
+   order and a positive integer.
+
+   Write a script to find out the number of unique Arithmetic Triplets
+   satisfying the following rules:
+a) i < j < k
+b) nums[j] - nums[i] == diff
+c) nums[k] - nums[j] == diff
+
+Example 1
+
+Input: @nums = (0, 1, 4, 6, 7, 10)
+       $diff = 3
+Output: 2
+
+Index (1, 2, 4) is an arithmetic triplet because both  7 - 4 == 3 and 4 - 1 == 3
+.
+Index (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3
+.
+
+Example 2
+
+Input: @nums = (4, 5, 6, 7, 8, 9)
+       $diff = 2
+Output: 2
+
+(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.
+(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.
+
+Task 2: Prime Order
+"""
+### solution by pokgopun@gmail.com
+
+from itertools import combinations
+
+def cntArmtcTrplt(nums: tuple, diff: int):
+    return len(
+            tuple(
+                filter(lambda x: x[1] - x[0] == diff and 2*x[1] == x[0] + x[2], combinations(nums,3))
+                )
+            )
+
+for (num, diff), count in {
+        ((0, 1, 4, 6, 7, 10), 3): 2,
+        ((4, 5, 6, 7, 8, 9), 2): 2,
+        }.items():
+    print(cntArmtcTrplt(num,diff)==count)
diff --git a/challenge-241/pokgopun/python/ch-2.py b/challenge-241/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-241/
+"""
+
+Task 2: Prime Order
+
+Submitted by: [43]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of unique positive integers greater than 2.
+
+   Write a script to sort them in ascending order of the count of their
+   prime factors, tie-breaking by ascending value.
+
+Example 1
+
+Input: @int = (11, 8, 27, 4)
+Output: (11, 4, 8, 27))
+
+Prime factors of 11 => 11
+Prime factors of  4 => 2, 2
+Prime factors of  8 => 2, 2, 2
+Prime factors of 27 => 3, 3, 3
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 5th November
+   2023.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+from sympy.ntheory import factorint
+
+def sortByFactorCount(tup: tuple):
+        return tuple(
+                sorted(
+                    sorted(tup), key = lambda x: sum(factorint(x).values())
+                    )
+                )
+
+for inpt,otpt in {
+        (11, 8, 27, 4): (11, 4, 8, 27),
+        (11, 27, 8, 4): (11, 4, 8, 27),
+        }.items():
+    print(sortByFactorCount(inpt)==otpt)
+