pwc285 solution in python

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diff --git a/challenge-285/pokgopun/python/ch-1.py b/challenge-285/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-285/
+"""
+
+Task 1: No Connection
+
+Submitted by: [49]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given a list of routes, @routes.
+
+   Write a script to find the destination with no further outgoing
+   connection.
+
+Example 1
+
+Input: @routes = (["B","C"], ["D","B"], ["C","A"])
+Output: "A"
+
+"D" -> "B" -> "C" -> "A".
+"B" -> "C" -> "A".
+"C" -> "A".
+"A".
+
+Example 2
+
+Input: @routes = (["A","Z"])
+Output: "Z"
+
+Task 2: Making Change
+"""
+### solution by pokgopun@gmail.com
+
+def nc(routes: tuple):
+    srcs = tuple(e[0] for e in routes)
+    for src, dst in routes:
+        if dst not in srcs:
+            return dst
+    return None
+
+import unittest
+
+class TestNc(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                (("B","C"), ("D","B"), ("C","A")): "A",
+                (("A","Z"),): "Z",
+                }.items():
+            self.assertEqual(nc(inpt),otpt)
+
+unittest.main()
diff --git a/challenge-285/pokgopun/python/ch-2.py b/challenge-285/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-285/
+"""
+
+Task 2: Making Change
+
+Submitted by: [50]David Ferrone
+     __________________________________________________________________
+
+   Compute the number of ways to make change for given amount in cents. By
+   using the coins e.g. Penny, Nickel, Dime, Quarter and Half-dollar, in
+   how many distinct ways can the total value equal to the given amount?
+   Order of coin selection does not matter.
+A penny (P) is equal to 1 cent.
+A nickel (N) is equal to 5 cents.
+A dime (D) is equal to 10 cents.
+A quarter (Q) is equal to 25 cents.
+A half-dollar (HD) is equal to 50 cents.
+
+Example 1
+
+Input: $amount = 9
+Ouput: 2
+
+1: 9P
+2: N + 4P
+
+Example 2
+
+Input: $amount = 15
+Ouput: 6
+
+1: D + 5P
+2: D + N
+3: 3N
+4: 2N + 5P
+5: N + 10P
+6: 15P
+
+Example 3
+
+Input: $amount = 100
+Ouput: 292
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 8th September
+   2024.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def mc(amount: int):
+    c = 0
+    for a50 in range(0,amount+1,50):
+        for a25 in range(0,amount-a50+1,25):
+            for a10 in range(0,amount-a50-a25+1,10):
+                for a5 in range(0,amount-a50-a25-a10+1,5):
+                    for a1 in range(0,amount-a50-a25-a10-a5+1,1):
+                        if a50 + a25 + a10 + a5 + a1 == amount:
+                            c += 1
+    return c
+
+import unittest
+
+class TestMc(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                9: 2,
+                15: 6,
+                100: 292,
+                }.items():
+            self.assertEqual(mc(inpt),otpt)
+
+unittest.main()