Merge pull request #9116 from pokgopun/pwc244

pwc244 solution in go

2 files changed, 170 insertions, 0 deletions

diff --git a/challenge-244/pokgopun/go/ch-1.go b/challenge-244/pokgopun/go/ch-1.go
new file mode 100644
index 0000000000..8a781161fb
--- /dev/null
+++ b/challenge-244/pokgopun/go/ch-1.go
@@ -0,0 +1,71 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-244/
+/*#
+
+Task 1: Count Smaller
+
+Submitted by: [45]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers.
+
+   Write a script to calculate the number of integers smaller than the
+   integer at each index.
+
+Example 1
+
+Input: @int = (8, 1, 2, 2, 3)
+Output: (4, 0, 1, 1, 3)
+
+For index = 0, count of elements less 8 is 4.
+For index = 1, count of elements less 1 is 0.
+For index = 2, count of elements less 2 is 1.
+For index = 3, count of elements less 2 is 1.
+For index = 4, count of elements less 3 is 3.
+
+Example 2
+
+Input: @int = (6, 5, 4, 8)
+Output: (2, 1, 0, 3)
+
+Example 3
+
+Input: @int = (2, 2, 2)
+Output: (0, 0, 0)
+
+Task 2: Group Hero
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+	"fmt"
+	"reflect"
+)
+
+type mArr []int
+
+func (ma mArr) conv() (r mArr) {
+	l := len(ma)
+	r = make(mArr, l)
+	for i, v := range ma {
+		for j := 0; j < l; j++ {
+			if ma[j] < v {
+				r[i]++
+			}
+		}
+	}
+	return r
+}
+
+func main() {
+	for _, data := range []struct {
+		input, output mArr
+	}{
+		{mArr{8, 1, 2, 2, 3}, mArr{4, 0, 1, 1, 3}},
+		{mArr{6, 5, 4, 8}, mArr{2, 1, 0, 3}},
+		{mArr{2, 2, 2}, mArr{0, 0, 0}},
+	} {
+		fmt.Println(reflect.DeepEqual(data.input.conv(), data.output))
+	}
+}
diff --git a/challenge-244/pokgopun/go/ch-2.go b/challenge-244/pokgopun/go/ch-2.go
new file mode 100644
index 0000000000..69b3a4c228
--- /dev/null
+++ b/challenge-244/pokgopun/go/ch-2.go
@@ -0,0 +1,99 @@
+//# https://theweeklychallenge.org/blog/perl-weekly-challenge-244/
+/*#
+
+Task 2: Group Hero
+
+Submitted by: [46]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers representing the strength.
+
+   Write a script to return the sum of the powers of all possible
+   combinations; power is defined as the square of the largest number in a
+   sequence, multiplied by the smallest.
+
+Example 1
+
+Input: @nums = (2, 1, 4)
+Output: 141
+
+Group 1: (2) => square(max(2)) * min(2) => 4 * 2 => 8
+Group 2: (1) => square(max(1)) * min(1) => 1 * 1 => 1
+Group 3: (4) => square(max(4)) * min(4) => 16 * 4 => 64
+Group 4: (2,1) => square(max(2,1)) * min(2,1) => 4 * 1 => 4
+Group 5: (2,4) => square(max(2,4)) * min(2,4) => 16 * 2 => 32
+Group 6: (1,4) => square(max(1,4)) * min(1,4) => 16 * 1 => 16
+Group 7: (2,1,4) => square(max(2,1,4)) * min(2,1,4) => 16 * 1 => 16
+
+Sum: 8 + 1 + 64 + 4 + 32 + 16 + 16 => 141
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 26th November
+   2023.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+#*/
+//# solution by pokgopun@gmail.com
+
+package main
+
+import (
+	"fmt"
+)
+
+type mArr []int
+
+func (ma mArr) power() int {
+	mx := ma.m(true)
+	return mx * mx * ma.m(false)
+}
+
+func (ma mArr) m(b bool) (m int) {
+	l := len(ma)
+	f := func(x, y int) int {
+		if b {
+			return max(x, y)
+		}
+		return min(x, y)
+	}
+	if l > 0 {
+		m = ma[0]
+		for l > 1 {
+			l--
+			m = f(m, ma[l])
+		}
+	}
+	return m
+}
+
+func (ma mArr) sumOfPower() (s int) {
+	l := len(ma)
+	ma0 := make(mArr, l)
+	var d, i, j int
+	for n := 1; n < 1<<l; n++ {
+		d, i, j = n, 0, 0
+		for d > 0 {
+			if d%2 > 0 {
+				ma0[j] = ma[i]
+				j++
+			}
+			i++
+			d /= 2
+		}
+		s += ma0[:j].power()
+		clear(ma0)
+	}
+	return s
+}
+
+func main() {
+	for _, data := range []struct {
+		input  mArr
+		output int
+	}{
+		{mArr{2, 1, 4}, 141},
+	} {
+		fmt.Println(data.input.sumOfPower() == data.output)
+	}
+}