Merge pull request #12161 from pokgopun/pwc325

Pwc325

4 files changed, 397 insertions, 0 deletions

diff --git a/challenge-325/pokgopun/go/ch-1.go b/challenge-325/pokgopun/go/ch-1.go
new file mode 100644
index 0000000000..9ec1056d7a
--- /dev/null
+++ b/challenge-325/pokgopun/go/ch-1.go
@@ -0,0 +1,93 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-325/
+/*#
+
+Task 1: Consecutive One
+
+Submitted by: [43]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given a binary array containing only 0 or/and 1.
+
+   Write a script to find out the maximum consecutive 1 in the given
+   array.
+
+Example 1
+
+Input: @binary = (0, 1, 1, 0, 1, 1, 1)
+Output: 3
+
+Example 2
+
+Input: @binary = (0, 0, 0, 0)
+Output: 0
+
+Example 3
+
+Input: @binary = (1, 0, 1, 0, 1, 1)
+Output: 2
+
+Task 2: Final Price
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+	"io"
+	"os"
+
+	"github.com/google/go-cmp/cmp"
+)
+
+type ints []int
+
+func (in ints) process() int {
+	var c, mx int
+	n := len(in)
+	for n > 0 {
+		n--
+		if in[n] == 0 {
+			if c == 0 {
+				continue
+			}
+			if c > mx {
+				//fmt.Printf("mx = %d\n", c)
+				mx = c
+			}
+			if n > 0 && n <= mx {
+				//fmt.Printf("remaining %d cannot exceed max %d\n", n, mx)
+				break
+			}
+			c = 0
+		} else {
+			c++
+		}
+	}
+	if n == 0 && in[n] > 0 {
+		//fmt.Println("final max update")
+		if mx < c {
+			//fmt.Printf("mx = %d\n", c)
+			mx = c
+		}
+	}
+	return mx
+}
+
+func main() {
+	for _, data := range []struct {
+		input  ints
+		output int
+	}{
+		{ints{0, 1, 1, 0, 1, 1, 1}, 3},
+		{ints{0, 0, 0, 0}, 0},
+		{ints{1, 0, 1, 0, 1, 1}, 2},
+		{ints{1, 0, 1, 1, 0, 1, 1}, 2},
+		{ints{1, 0, 1, 1, 0, 0, 1, 1}, 2},
+		{ints{0, 1, 1, 1, 0, 1, 1, 1}, 3},
+		{ints{1, 1, 1, 1, 0, 1, 1, 1, 0}, 4},
+		{ints{0, 1, 1, 0, 0, 1, 1, 1, 0}, 3},
+	} {
+		//fmt.Println(data)
+		io.WriteString(os.Stdout, cmp.Diff(data.input.process(), data.output)) // blank if ok, otherwise show the difference
+	}
+}
diff --git a/challenge-325/pokgopun/go/ch-2.go b/challenge-325/pokgopun/go/ch-2.go
new file mode 100644
index 0000000000..830361216f
--- /dev/null
+++ b/challenge-325/pokgopun/go/ch-2.go
@@ -0,0 +1,122 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-325/
+/*#
+
+Task 2: Final Price
+
+Submitted by: [44]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given an array of item prices.
+
+   Write a script to find out the final price of each items in the given
+   array.
+
+   There is a special discount scheme going on. If there’s an item with a
+   lower or equal price later in the list, you get a discount equal to
+   that later price (the first one you find in order).
+
+Example 1
+
+Input: @prices = (8, 4, 6, 2, 3)
+Output: (4, 2, 4, 2, 3)
+
+Item 0:
+The item price is 8.
+The first time that has price <= current item price is 4.
+Final price = 8 - 4 => 4
+
+Item 1:
+The item price is 4.
+The first time that has price <= current item price is 2.
+Final price = 4 - 2 => 2
+
+Item 2:
+The item price is 6.
+The first time that has price <= current item price is 2.
+Final price = 6 - 2 => 4
+
+Item 3:
+The item price is 2.
+No item has price <= current item price, no discount.
+Final price = 2
+
+Item 4:
+The item price is 3.
+Since it is the last item, so no discount.
+Final price = 3
+
+Example 2
+
+Input: @prices = (1, 2, 3, 4, 5)
+Output: (1, 2, 3, 4, 5)
+
+Example 3
+
+Input: @prices = (7, 1, 1, 5)
+Output: (6, 0, 1, 5)
+
+Item 0:
+The item price is 7.
+The first time that has price <= current item price is 1.
+Final price = 7 - 1 => 6
+
+Item 1:
+The item price is 1.
+The first time that has price <= current item price is 1.
+Final price = 1 - 1 => 0
+
+Item 2:
+The item price is 1.
+No item has price <= current item price, so no discount.
+Final price = 1
+
+Item 3:
+The item price is 5.
+Since it is the last item, so no discount.
+Final price = 5
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 15th June 2025.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+	"io"
+	"os"
+	"slices"
+
+	"github.com/google/go-cmp/cmp"
+)
+
+type ints []int
+
+func (in ints) process() ints {
+	out := slices.Clone(in)
+	l := len(out)
+	for i := range l - 1 {
+		for _, v := range out[i+1:] {
+			if out[i] >= v {
+				out[i] -= v
+				break
+			}
+		}
+	}
+	return out
+}
+
+func main() {
+	for _, data := range []struct {
+		input, output ints
+	}{
+		{ints{8, 4, 6, 2, 3}, ints{4, 2, 4, 2, 3}},
+		{ints{1, 2, 3, 4, 5}, ints{1, 2, 3, 4, 5}},
+		{ints{7, 1, 1, 5}, ints{6, 0, 1, 5}},
+	} {
+		io.WriteString(os.Stdout, cmp.Diff(data.input.process(), data.output)) // blank if ok, otherwise show the difference
+	}
+}
diff --git a/challenge-325/pokgopun/python/ch-1.py b/challenge-325/pokgopun/python/ch-1.py
new file mode 100644
index 0000000000..6b0cc53274
--- /dev/null
+++ b/challenge-325/pokgopun/python/ch-1.py
@@ -0,0 +1,76 @@
+### https://theweeklychallenge.org/blog/perl-weekly-challenge-325/
+"""
+
+Task 1: Consecutive One
+
+Submitted by: [43]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given a binary array containing only 0 or/and 1.
+
+   Write a script to find out the maximum consecutive 1 in the given
+   array.
+
+Example 1
+
+Input: @binary = (0, 1, 1, 0, 1, 1, 1)
+Output: 3
+
+Example 2
+
+Input: @binary = (0, 0, 0, 0)
+Output: 0
+
+Example 3
+
+Input: @binary = (1, 0, 1, 0, 1, 1)
+Output: 2
+
+Task 2: Final Price
+"""
+### solution by pokgopun@gmail.com
+
+def co(bins: tuple[int]) -> int:
+    c, mx = 0, 0
+    n = len(bins)
+    while n > 0:
+        n -= 1
+        if bins[n] == 0:
+            if c == 0:
+                continue
+            if c > mx:
+                #print("mx =",c)
+                mx = c
+            if n > 0 and n <= mx:
+                #print(f'remaning {n} cannot make up mx')
+                break
+            c = 0
+        else:
+            c += 1
+    if n ==0 and bins[n] > 0:
+        #print("final mx update")
+        if c > mx:
+            #print("mx =",c)
+            mx = c
+    return mx
+
+import unittest
+
+class TestCo(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                (0, 1, 1, 0, 1, 1, 1): 3,
+                (0, 0, 0, 0): 0,
+                (1, 0, 1, 0, 1, 1): 2,
+                (1, 0, 1, 1, 0, 1, 1): 2,
+                (1, 0, 1, 1, 0, 0, 1, 1): 2,
+                (0, 1, 1, 1, 0, 1, 1, 1): 3,
+                (0, 1, 1, 1, 0, 1, 1, 1,0): 3,
+                (0, 1, 1, 0, 0, 1, 1, 1,0): 3,
+                (1, 1, 1, 0, 0, 1, 1, 1,0): 3,
+                (1, 1, 1, 1, 0, 1, 1, 1,0): 4,
+                }.items():
+            #print(inpt,otpt)
+            self.assertEqual(co(inpt),otpt)
+
+unittest.main()
diff --git a/challenge-325/pokgopun/python/ch-2.py b/challenge-325/pokgopun/python/ch-2.py
new file mode 100644
index 0000000000..cd9c183f19
--- /dev/null
+++ b/challenge-325/pokgopun/python/ch-2.py
@@ -0,0 +1,106 @@
+### https://theweeklychallenge.org/blog/perl-weekly-challenge-325/
+"""
+
+Task 2: Final Price
+
+Submitted by: [44]Mohammad Sajid Anwar
+     __________________________________________________________________
+
+   You are given an array of item prices.
+
+   Write a script to find out the final price of each items in the given
+   array.
+
+   There is a special discount scheme going on. If there’s an item with a
+   lower or equal price later in the list, you get a discount equal to
+   that later price (the first one you find in order).
+
+Example 1
+
+Input: @prices = (8, 4, 6, 2, 3)
+Output: (4, 2, 4, 2, 3)
+
+Item 0:
+The item price is 8.
+The first time that has price <= current item price is 4.
+Final price = 8 - 4 => 4
+
+Item 1:
+The item price is 4.
+The first time that has price <= current item price is 2.
+Final price = 4 - 2 => 2
+
+Item 2:
+The item price is 6.
+The first time that has price <= current item price is 2.
+Final price = 6 - 2 => 4
+
+Item 3:
+The item price is 2.
+No item has price <= current item price, no discount.
+Final price = 2
+
+Item 4:
+The item price is 3.
+Since it is the last item, so no discount.
+Final price = 3
+
+Example 2
+
+Input: @prices = (1, 2, 3, 4, 5)
+Output: (1, 2, 3, 4, 5)
+
+Example 3
+
+Input: @prices = (7, 1, 1, 5)
+Output: (6, 0, 1, 5)
+
+Item 0:
+The item price is 7.
+The first time that has price <= current item price is 1.
+Final price = 7 - 1 => 6
+
+Item 1:
+The item price is 1.
+The first time that has price <= current item price is 1.
+Final price = 1 - 1 => 0
+
+Item 2:
+The item price is 1.
+No item has price <= current item price, so no discount.
+Final price = 1
+
+Item 3:
+The item price is 5.
+Since it is the last item, so no discount.
+Final price = 5
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 15th June 2025.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def fp(ints: tuple[int]) -> tuple[int]:
+    lst = list(ints)
+    for i in range(len(lst)-1):
+        for v in lst[i+1:]:
+            if lst[i] >= v:
+                lst[i] -= v
+                break
+    return tuple(lst)
+
+import unittest
+
+class TestFp(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                (8, 4, 6, 2, 3): (4, 2, 4, 2, 3),
+                (1, 2, 3, 4, 5): (1, 2, 3, 4, 5),
+                (7, 1, 1, 5): (6, 0, 1, 5),
+                }.items():
+            self.assertEqual(fp(inpt),otpt)
+
+unittest.main()