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diff --git a/challenge-249/pokgopun/python/ch-1.py b/challenge-249/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-249/
+"""
+
+Task 1: Shortest Distance
+
+Submitted by: [66]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers with even number of elements.
+
+   Write a script to divide the given array into equal pairs such that:
+a) Each element belongs to exactly one pair.
+b) The elements present in a pair are equal.
+
+Example 1
+
+Input: @ints = (3, 2, 3, 2, 2, 2)
+Output: (2, 2), (3, 3), (2, 2)
+
+There are 6 elements in @ints.
+They should be divided into 6 / 2 = 3 pairs.
+@ints is divided into the pairs (2, 2), (3, 3), and (2, 2) satisfying all the co
+nditions.
+
+Example 2
+
+Input: @ints = (1, 2, 3, 4)
+Output: ()
+
+There is no way to divide @ints 2 pairs such that the pairs satisfy every condit
+ion.
+
+Task 2: DI String Match
+"""
+### solution by pokgopun@gmail.com
+
+from itertools import chain
+
+def equalPairs(tup: tuple):
+    return tuple(
+            chain.from_iterable(
+                ( (e,e) for c in range(tup.count(e)//2 ) )
+                for e in set(tup)
+                )
+            )
+
+import unittest
+
+class TestEqualPairs(unittest.TestCase):
+    def test1(self):
+        for inpt,otpt in {
+                (3, 2, 3, 2, 2, 2): ((2, 2), (3, 3), (2, 2)),
+                (1, 2, 3, 4): (),
+                }.items():
+            self.assertEqual(sorted(equalPairs(inpt)),sorted(otpt))
+
+unittest.main()
diff --git a/challenge-249/pokgopun/python/ch-2.py b/challenge-249/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-249/
+"""
+
+Task 2: DI String Match
+
+Submitted by: [67]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given a string s, consisting of only the characters "D" and
+   "I".
+
+   Find a permutation of the integers [0 .. length(s)] such that for each
+   character s[i] in the string:
+s[i] == 'I' ⇒ perm[i] < perm[i + 1]
+s[i] == 'D' ⇒ perm[i] > perm[i + 1]
+
+Example 1
+
+Input: $str = "IDID"
+Output: (0, 4, 1, 3, 2)
+
+Example 2
+
+Input: $str = "III"
+Output: (0, 1, 2, 3)
+
+Example 3
+
+Input: $str = "DDI"
+Output: (3, 2, 0, 1)
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 31st December
+   2023.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+from itertools import permutations
+
+def dism(s: str):
+    t = tuple()
+    l = len(s)
+    for perm in permutations(tuple(range(l+1)),l+1):
+        #print(perm)
+        for i in range(l):
+            #print(f"{s[i]},{perm[i]},{perm[i+1]}")
+            if s[i]=="I":
+                if perm[i] >= perm[i+1]:
+                    break
+            elif s[i]=="D":
+                if perm[i] <= perm[i+1]:
+                    break
+        else:
+            #print("=>",perm)
+            t += (perm,)
+    #print(t)
+    return t
+
+import unittest
+
+class TestDism(unittest.TestCase):
+    def test(self):
+        for inpt,otpt in {
+                "IDID": (0, 4, 1, 3, 2),
+                "III": (0, 1, 2, 3),
+                "DDI": (3, 2, 0, 1),
+                }.items():
+            self.assertEqual(dism(inpt).count(otpt)>0,True)
+
+unittest.main()