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-rw-r--r--challenge-271/pokgopun/go/ch-1.go124
-rw-r--r--challenge-271/pokgopun/go/ch-2.go87
-rw-r--r--challenge-271/pokgopun/python/ch-1.py69
-rw-r--r--challenge-271/pokgopun/python/ch-2.py72
4 files changed, 352 insertions, 0 deletions
diff --git a/challenge-271/pokgopun/go/ch-1.go b/challenge-271/pokgopun/go/ch-1.go
new file mode 100644
index 0000000000..d2e9c3e0ba
--- /dev/null
+++ b/challenge-271/pokgopun/go/ch-1.go
@@ -0,0 +1,124 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-271/
+/*#
+
+Task 1: Maximum Ones
+
+Submitted by: [42]Mohammad Sajid Anwar
+ __________________________________________________________________
+
+ You are given a m x n binary matrix.
+
+ Write a script to return the row number containing maximum ones, in
+ case of more than one rows then return smallest row number.
+
+Example 1
+
+Input: $matrix = [ [0, 1],
+ [1, 0],
+ ]
+Output: 1
+
+Row 1 and Row 2 have the same number of ones, so return row 1.
+
+Example 2
+
+Input: $matrix = [ [0, 0, 0],
+ [1, 0, 1],
+ ]
+Output: 2
+
+Row 2 has the maximum ones, so return row 2.
+
+Example 3
+
+Input: $matrix = [ [0, 0],
+ [1, 1],
+ [0, 0],
+ ]
+Output: 2
+
+Row 2 have the maximum ones, so return row 2.
+
+Task 2: Sort by 1 bits
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+ "io"
+ "os"
+
+ "github.com/google/go-cmp/cmp"
+)
+
+type answer struct {
+ c, i int
+}
+
+type row []int
+
+func (rw row) countOne() int {
+ c := 0
+ l := len(rw)
+ for i := 0; i < l; i++ {
+ if rw[i] == 1 {
+ c++
+ }
+ }
+ return c
+}
+
+type matrix []row
+
+func (mtx matrix) maxOneRow() int {
+ mx := len(mtx[0])
+ c := mtx[0].countOne()
+ if c == mx {
+ return 1
+ }
+ ans := answer{c: c}
+ l := len(mtx)
+ for i := 1; i < l; i++ {
+ c = mtx[i].countOne()
+ if c == mx {
+ return i + 1
+ }
+ if c > ans.c {
+ ans = answer{c, i}
+ }
+ }
+ return ans.i + 1
+}
+
+func main() {
+ for _, data := range []struct {
+ input matrix
+ output int
+ }{
+ {
+ matrix{
+ row{0, 1},
+ row{1, 0},
+ },
+ 1,
+ },
+ {
+ matrix{
+ row{0, 0, 0},
+ row{1, 0, 1},
+ },
+ 2,
+ },
+ {
+ matrix{
+ row{0, 0},
+ row{1, 1},
+ row{0, 0},
+ },
+ 2,
+ },
+ } {
+ io.WriteString(os.Stdout, cmp.Diff(data.input.maxOneRow(), data.output)) // blank if ok, otherwise show the difference
+ }
+}
diff --git a/challenge-271/pokgopun/go/ch-2.go b/challenge-271/pokgopun/go/ch-2.go
new file mode 100644
index 0000000000..9f109be280
--- /dev/null
+++ b/challenge-271/pokgopun/go/ch-2.go
@@ -0,0 +1,87 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-271/
+/*#
+
+Task 2: Sort by 1 bits
+
+Submitted by: [43]Mohammad Sajid Anwar
+ __________________________________________________________________
+
+ You are give an array of integers, @ints.
+
+ Write a script to sort the integers in ascending order by the number of
+ 1 bits in their binary representation. In case more than one integers
+ have the same number of 1 bits then sort them in ascending order.
+
+Example 1
+
+Input: @ints = (0, 1, 2, 3, 4, 5, 6, 7, 8)
+Output: (0, 1, 2, 4, 8, 3, 5, 6, 7)
+
+0 = 0 one bits
+1 = 1 one bits
+2 = 1 one bits
+4 = 1 one bits
+8 = 1 one bits
+3 = 2 one bits
+5 = 2 one bits
+6 = 2 one bits
+7 = 3 one bits
+
+Example 2
+
+Input: @ints = (1024, 512, 256, 128, 64)
+Output: (64, 128, 256, 512, 1024)
+
+All integers in the given array have one 1-bits, so just sort them in ascending
+order.
+ __________________________________________________________________
+
+ Last date to submit the solution 23:59 (UK Time) Sunday 2nd June 2024.
+ __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+ "io"
+ "os"
+ "sort"
+
+ "github.com/google/go-cmp/cmp"
+)
+
+func countBitOn(n uint) uint {
+ var c uint = 0
+ for n > 0 {
+ if n%2 != 0 {
+ c++
+ }
+ n /= 2
+ }
+ return c
+}
+
+func sortBitOn(uints []uint) []uint {
+ sort.Slice(uints, func(i, j int) bool {
+ a, b := countBitOn(uints[i]), countBitOn(uints[j])
+ if a == b {
+ return uints[i] < uints[j]
+ }
+ return a < b
+ })
+ return uints
+}
+
+func main() {
+ for _, data := range []struct {
+ input, output []uint
+ }{
+ {[]uint{0, 1, 2, 3, 4, 5, 6, 7, 8}, []uint{0, 1, 2, 4, 8, 3, 5, 6, 7}},
+ {[]uint{1024, 512, 256, 128, 64}, []uint{64, 128, 256, 512, 1024}},
+ } {
+ io.WriteString(os.Stdout, cmp.Diff(sortBitOn(data.input), data.output)) // blank if ok, otherwise show the difference
+ }
+}
diff --git a/challenge-271/pokgopun/python/ch-1.py b/challenge-271/pokgopun/python/ch-1.py
new file mode 100644
index 0000000000..5e2b758e0a
--- /dev/null
+++ b/challenge-271/pokgopun/python/ch-1.py
@@ -0,0 +1,69 @@
+### https://theweeklychallenge.org/blog/perl-weekly-challenge-271/
+"""
+
+Task 1: Maximum Ones
+
+Submitted by: [42]Mohammad Sajid Anwar
+ __________________________________________________________________
+
+ You are given a m x n binary matrix.
+
+ Write a script to return the row number containing maximum ones, in
+ case of more than one rows then return smallest row number.
+
+Example 1
+
+input: $matrix = [ [0, 1],
+ [1, 0],
+ ]
+output: 1
+
+row 1 and row 2 have the same number of ones, so return row 1.
+
+example 2
+
+input: $matrix = [ [0, 0, 0],
+ [1, 0, 1],
+ ]
+output: 2
+
+row 2 has the maximum ones, so return row 2.
+
+example 3
+
+input: $matrix = [ [0, 0],
+ [1, 1],
+ [0, 0],
+ ]
+Output: 2
+
+Row 2 have the maximum ones, so return row 2.
+
+Task 2: Sort by 1 bits
+"""
+### solution by pokgopun@gmail.com
+
+def maxOne(mtx: list):
+ return 1 - max(
+ (mtx[e].count(1),-e) for e in range(len(mtx))
+ )[1]
+
+import unittest
+
+class TestMaxOne(unittest.TestCase):
+ def test(self):
+ for otpt, inpt in {
+ 1: [ [0, 1],
+ [1, 0],
+ ],
+ 2: [ [0, 0, 0],
+ [1, 0, 1],
+ ],
+ 2: [ [0, 0],
+ [1, 1],
+ [0, 0],
+ ],
+ }.items():
+ self.assertEqual(maxOne(inpt),otpt)
+
+unittest.main()
diff --git a/challenge-271/pokgopun/python/ch-2.py b/challenge-271/pokgopun/python/ch-2.py
new file mode 100644
index 0000000000..80ae73080e
--- /dev/null
+++ b/challenge-271/pokgopun/python/ch-2.py
@@ -0,0 +1,72 @@
+### https://theweeklychallenge.org/blog/perl-weekly-challenge-271/
+"""
+
+Task 2: Sort by 1 bits
+
+Submitted by: [43]Mohammad Sajid Anwar
+ __________________________________________________________________
+
+ You are give an array of integers, @ints.
+
+ Write a script to sort the integers in ascending order by the number of
+ 1 bits in their binary representation. In case more than one integers
+ have the same number of 1 bits then sort them in ascending order.
+
+Example 1
+
+Input: @ints = (0, 1, 2, 3, 4, 5, 6, 7, 8)
+Output: (0, 1, 2, 4, 8, 3, 5, 6, 7)
+
+0 = 0 one bits
+1 = 1 one bits
+2 = 1 one bits
+4 = 1 one bits
+8 = 1 one bits
+3 = 2 one bits
+5 = 2 one bits
+6 = 2 one bits
+7 = 3 one bits
+
+Example 2
+
+Input: @ints = (1024, 512, 256, 128, 64)
+Output: (64, 128, 256, 512, 1024)
+
+All integers in the given array have one 1-bits, so just sort them in ascending
+order.
+ __________________________________________________________________
+
+ Last date to submit the solution 23:59 (UK Time) Sunday 2nd June 2024.
+ __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def countBitOn(n):
+ c = 0
+ while n > 0:
+ if n % 2 > 0:
+ c += 1
+ n = n // 2
+ return c
+
+def sortBitOn(nums: tuple):
+ return tuple(
+ sorted(
+ nums,
+ key = lambda x: (countBitOn(x), x)
+ )
+ )
+
+import unittest
+
+class TestSortBitOn(unittest.TestCase):
+ def test(self):
+ for inpt, otpt in {
+ (0, 1, 2, 3, 4, 5, 6, 7, 8): (0, 1, 2, 4, 8, 3, 5, 6, 7),
+ (1024, 512, 256, 128, 64): (64, 128, 256, 512, 1024),
+ }.items():
+ self.assertEqual(sortBitOn(inpt),otpt)
+
+unittest.main()
generated by cgit (git 2.34.1) at 2026-07-01 16:04:39 +0000