I have a Go Language question here, is there any much better way to answer the answer in coding Golang compare to mine below?
Mangkuk is a list consisting of maximal size Sudu. Sudu is a permutation of consecutive integers, possibly with repeated items.
A Cawan is a Mangkuk where each Sudu is sorted in the ascending order. Write a function, MakeCawan(→Mangkuk), to sort the given Mangkuk into a Cawan.
For example,
MakeCawan([21, 20, 18, 20, 18, 20, 19]),
MakeCawan([21, 2000000, 18, 20, 18, 20, 19]),
MakeCawan([21, 20, 18, 20, 18, 20, 1900000])
should produce, respectively,
[18, 18, 19, 20, 20, 20, 21],
[21, 2000000, 18, 18, 19, 20, 20],
[20, 21, 18, 20, 18, 20, 1900000].
package main
import (
"fmt"
"sort"
)
func main() {
sl := []string{"MakeCawan"}
sort.Sort(sort.StringSlice(sl))
fmt.Println(sl)
sl1 := []string{"MakeCawan"}
sort.Sort(sort.StringSlice(sl1))
fmt.Println(sl1)
sl2 := []string{"MakeCawan"}
sort.Sort(sort.StringSlice(sl2))
fmt.Println(sl2)
intSlice := []int{21,20,18,20,18,20,19}
sort.Sort(sort.IntSlice(intSlice))
fmt.Println(intSlice)
}
The Output:
https://play.golang.org/p/tsE0BtMRos_9
</div>
The problem is a little bit tricky: It does not ask you to sort the whole slice (or mangkuk in its own term); it asks you to first recognize all successive intervals (with possible repeated elements) which is called sudu, and then sort each sudu.
func makeCawan(mangkuk []int) []int {
for now, n := 0, len(mangkuk); now < n; {
min := mangkuk[now]
max := min
head := now
loop:
for now++; now < n; now++ {
switch x := mangkuk[now]; {
case x < min-1 || x > max+1:
sort(mangkuk[head:now], min, max)
break loop
case x == min-1:
min = x
case x == max+1:
max = x
}
}
if now >= n {
sort(mangkuk[head:now], min, max)
}
}
return mangkuk
}
Playground: https://play.golang.org/p/z3TGWnWnrVY
Answering this question assuming you want to sort a slice of int which are consecutive and repetitive.
Simply sorting the slice is a readable solution but would take O(nlgn) using a comparsion based sorting algorithm for a slice of length n.
We can use a better performin algorithm with O(n) auxillary space. Algorithm:
1. Iterate over array A and find min and max in.
2. Create an array B with length max-min+1.
3. Iterate over A and store count of each element in B i.e B[A[i] - min]++
.
4. Now iterate over B and print i + min
, B[i] times.
Time Complexity - O(n)
https://play.golang.org/p/rptgMpWdKCX
Note that this loop is also O(n) where n is length of actual input array.
for i:=0;i<len(b);i++{
for b[i] != 0{
fmt.Printf("%v ", i + min)
b[i]--
}
}