pwc237 solution in python

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diff --git a/challenge-237/pokgopun/python/ch-1.py b/challenge-237/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-237/
+"""
+
+Task 1: Seize The Day
+
+Submitted by: [44]Mark Anderson
+     __________________________________________________________________
+
+   Given a year, a month, a weekday of month, and a day of week (1 (Mon)
+   .. 7 (Sun)), print the day.
+
+Example 1
+
+Input: Year = 2024, Month = 4, Weekday of month = 3, day of week = 2
+Output: 16
+
+The 3rd Tue of Apr 2024 is the 16th
+
+Example 2
+
+Input: Year = 2025, Month = 10, Weekday of month = 2, day of week = 4
+Output: 9
+
+The 2nd Thu of Oct 2025 is the 9th
+
+Example 3
+
+Input: Year = 2026, Month = 8, Weekday of month = 5, day of week = 3
+Output: 0
+
+There isn't a 5th Wed in Aug 2026
+
+Task 2: Maximise Greatness
+"""
+### solution by pokgopun@gmail.com
+
+import datetime
+
+def seizeTheDay(tup):
+    y,m,mw,wd = tup
+    dt = datetime.date(y, m, 1)
+    wdd = wd - dt.isoweekday()
+    if wdd < 0: wdd += 7
+    try:
+        return dt.replace(day = 1 + 7*(mw - 1) + wdd).day
+    except:
+        return 0
+
+for inpt,otpt in {
+        (2024,4,3,2): 16,
+        (2025,10,2,4): 9,
+        (2026,8,5,3): 0,
+        }.items():
+    print(seizeTheDay(inpt)==otpt)
+    
diff --git a/challenge-237/pokgopun/python/ch-2.py b/challenge-237/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-237/
+"""
+
+Task 2: Maximise Greatness
+
+Submitted by: [45]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers.
+
+   Write a script to permute the give array such that you get the maximum
+   possible greatness.
+
+     To determine greatness, nums[i] < perm[i] where 0 <= i < nums.length
+
+Example 1
+
+Input: @nums = (1, 3, 5, 2, 1, 3, 1)
+Output: 4
+
+One possible permutation: (2, 5, 1, 3, 3, 1, 1) which returns 4 greatness as bel
+ow:
+nums[0] < perm[0]
+nums[1] < perm[1]
+nums[3] < perm[3]
+nums[4] < perm[4]
+
+Example 2
+
+Input: @ints = (1, 2, 3, 4)
+Output: 3
+
+One possible permutation: (2, 3, 4, 1) which returns 3 greatness as below:
+nums[0] < perm[0]
+nums[1] < perm[1]
+nums[2] < perm[2]
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 8th October
+   2023.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+from itertools import permutations
+
+def maxGreat(tup):
+    l, m = len(tup), 0
+    for pm in permutations(tup):
+        g = tuple( pm[i] > tup[i] for i in range(l) ).count(True)
+        if m < g : m = g
+    return m
+
+for inpt,otpt in {
+        (1, 3, 5, 2, 1, 3, 1): 4,
+        (1, 2, 3, 4): 3,
+        }.items():
+    print(maxGreat(inpt)==otpt)
+
+