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diff --git a/challenge-282/pokgopun/python/ch-1.py b/challenge-282/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-282/
+"""
+
+Task 1: Good Integer
+
+Submitted by: [57]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given a positive integer, $int, having 3 or more digits.
+
+   Write a script to return the Good Integer in the given integer or -1 if
+   none found.
+
+     A good integer is exactly three consecutive matching digits.
+
+Example 1
+
+Input: $int = 12344456
+Output: "444"
+
+Example 2
+
+Input: $int = 1233334
+Output: -1
+
+Example 3
+
+Input: $int = 10020003
+Output: "000"
+
+Task 2: Changing Keys
+"""
+### solution by pokgopun@gmail.com
+
+def gi(num: int):
+    count = 0
+    digit = ""
+    for d in str(num):
+        if digit == d:
+            count += 1
+        else:
+            if count == 3:
+                return digit*3
+            digit = d
+            count = 1
+    return -1
+
+import unittest
+
+class TestGi(unittest.TestCase):
+    def test(self):
+        for inpt,otpt in {
+                12344456: "444",
+                1233334: -1,
+                10020003: "000",
+                }.items():
+            self.assertEqual(gi(inpt),otpt)
+
+unittest.main()
diff --git a/challenge-282/pokgopun/python/ch-2.py b/challenge-282/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-282/
+"""
+
+Task 2: Changing Keys
+
+Submitted by: [58]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given an alphabetic string, $str, as typed by user.
+
+   Write a script to find the number of times user had to change the key
+   to type the given string. Changing key is defined as using a key
+   different from the last used key. The shift and caps lock keys won’t be
+   counted.
+
+Example 1
+
+Input: $str = 'pPeERrLl'
+Ouput: 3
+
+p -> P : 0 key change
+P -> e : 1 key change
+e -> E : 0 key change
+E -> R : 1 key change
+R -> r : 0 key change
+r -> L : 1 key change
+L -> l : 0 key change
+
+Example 2
+
+Input: $str = 'rRr'
+Ouput: 0
+
+Example 3
+
+Input: $str = 'GoO'
+Ouput: 1
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 18th August
+   2024.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def ck(string: str):
+    string = string.lower()
+    count = 0
+    for i in range(len(string)-1):
+        if string[i]!=string[i+1]:
+            count += 1
+    return count
+
+import unittest
+
+class TestCk(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                'pPeERrLl': 3,
+                'rRr': 0,
+                'GoO': 1,
+                }.items():
+            self.assertEqual(ck(inpt),otpt)
+
+unittest.main()
+