aboutsummaryrefslogtreecommitdiff
diff options
context:
space:
mode:
Diffstat
-rw-r--r--challenge-285/pokgopun/go/ch-1.go75
-rw-r--r--challenge-285/pokgopun/go/ch-2.go90
-rw-r--r--challenge-285/pokgopun/python/ch-1.py50
-rw-r--r--challenge-285/pokgopun/python/ch-2.py75
4 files changed, 290 insertions, 0 deletions
diff --git a/challenge-285/pokgopun/go/ch-1.go b/challenge-285/pokgopun/go/ch-1.go
new file mode 100644
index 0000000000..25e61b4565
--- /dev/null
+++ b/challenge-285/pokgopun/go/ch-1.go
@@ -0,0 +1,75 @@
+//# 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
+
+package main
+
+import (
+ "io"
+ "os"
+
+ "github.com/google/go-cmp/cmp"
+)
+
+type route struct {
+ src, dst string
+}
+
+type routes []route
+
+func (rs routes) nc() string {
+ m := make(map[string]bool)
+ for _, v := range rs {
+ m[v.src] = false
+ _, ok := m[v.dst]
+ if !ok {
+ m[v.dst] = true
+ }
+ }
+ for k, v := range m {
+ if v {
+ return k
+ }
+ }
+ return ""
+}
+
+func main() {
+ for _, data := range []struct {
+ input routes
+ output string
+ }{
+ {routes{route{"B", "C"}, route{"D", "B"}, route{"C", "A"}}, "A"},
+ {routes{route{"A", "Z"}}, "Z"},
+ } {
+ io.WriteString(os.Stdout, cmp.Diff(data.input.nc(), data.output)) // blank if ok, otherwise show the difference
+ }
+}
diff --git a/challenge-285/pokgopun/go/ch-2.go b/challenge-285/pokgopun/go/ch-2.go
new file mode 100644
index 0000000000..58a63ee266
--- /dev/null
+++ b/challenge-285/pokgopun/go/ch-2.go
@@ -0,0 +1,90 @@
+//# 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
+
+package main
+
+import (
+ "io"
+ "os"
+
+ "github.com/google/go-cmp/cmp"
+)
+
+func mc(a int) int {
+ c := 0
+ for c50 := range a/50 + 1 {
+ for c25 := range (a-50*c50)/25 + 1 {
+ for c10 := range (a-50*c50-25*c25)/10 + 1 {
+ for c5 := range (a-50*c50-25*c25-10*c10)/5 + 1 {
+ for c1 := range (a - 50*c50 - 25*c25 - 10*c10 - 5*c5) + 1 {
+ if 50*c50+25*c25+10*c10+5*c5+c1 == a {
+ c++
+ }
+ }
+ }
+ }
+ }
+ }
+ return c
+}
+
+func main() {
+ for _, data := range []struct {
+ input, output int
+ }{
+ {9, 2},
+ {15, 6},
+ {100, 292},
+ } {
+ io.WriteString(os.Stdout, cmp.Diff(mc(data.input), data.output)) // blank if ok, otherwise show the difference
+ }
+}
diff --git a/challenge-285/pokgopun/python/ch-1.py b/challenge-285/pokgopun/python/ch-1.py
new file mode 100644
index 0000000000..fcebb07e89
--- /dev/null
+++ b/challenge-285/pokgopun/python/ch-1.py
@@ -0,0 +1,50 @@
+### 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
new file mode 100644
index 0000000000..813b8cf378
--- /dev/null
+++ b/challenge-285/pokgopun/python/ch-2.py
@@ -0,0 +1,75 @@
+### 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()
generated by cgit (git 2.34.1) at 2026-07-04 03:05:20 +0000