Related
I'm currently writing a program and I want to randomly generate a matrix.
Currently I'm pre-setting the values in it as follows:
m1 := [3][3]int{
[3]int{1, 1, 1},
[3]int{4, 1, 7},
[3]int{1, 65, 1},
}
However I want the values inputted to be randomly generated in a range from 1-100.
import "math/rand"
I am importing the above library and trying to utilise it.
I have attempted to get this working however can't seem to make any headway.
m1 := [3][3]int{
[3]int{rand.Intn, 1, 1},
[3]int{4, 1, 7},
[3]int{1, 65, 1},
}
I have attempted to complete it with the above solution to make the first number random however I get the following error.
cannot use rand.Intn (type func(int) int) as type int in array or slice literal
Any help greatly appreciated.
The direct answer is the fact that rand.Intn() generates a random integer between 0 and n, where n is a parameter to this method. The error that you are getting is the compiler complaining that you are trying to initialize an int value with a function that requires two ints and returns one - you are trying to assign a function to an int. So the correct call would be something like rand.Intn(100), which will give you a random number between 0 - 100.
However, why do it this way? Why not dynamically initialize your array with random numbers as:
m1 := [3][3]int{}
for i:=0; i<3; i++ {
for j:=0; j<3; j++ {
m1[i][j] = rand.Int()
}
}
Answer to your question is answered above, this is an extension,
While rand.Int(10) always gives you 1, as it isn't seeded,
you can add this function to get random values each time you run your program,
package main
import (
"fmt"
"math/rand"
"time"
)
func init() {
rand.Seed(time.Now().UnixNano())
//we are seeding the rand variable with present time
//so that we would get different output each time
}
func main() {
randMatrix := make([][]int, 3)
// we have created a slice with length 3
//which can hold type []int, these can be of different length
for i := 0; i < 3; i++ {
randMatrix[i] = make([]int, 3)
// we are creating a slice which can hold type int
}
generate(randMatrix)
fmt.Println(randMatrix)
}
func generate(randMatrix [][]int) {
for i, innerArray := range randMatrix {
for j := range innerArray {
randMatrix[i][j] = rand.Intn(100)
//looping over each element of array and assigning it a random variable
}
}
}
This code generates random Matrix, below 100, while you can also use flags for any kind of future use and generalize the values,
import "flag"
var outerDim, innerDim, limit *int
func main() {
outerDim = flag.Int("outerDim", 3, "Outer dimension of the matrix")
innerDim = flag.Int("innerDim", 3, "inner dimenstion of the matrix")
limit = flag.Int("limit", 100, "matrix values are limited specified value")
flag.Parse()
randMatrix := make([][]int, *outerDim)
for i := 0; i < *outerDim; i++ {
randMatrix[i] = make([]int, *innerDim)
}
generate(randMatrix)
printMatrix(randMatrix)
}
func generate(randMatrix [][]int) {
for i, innerArray := range randMatrix {
for j := range innerArray {
randMatrix[i][j] = rand.Intn(*limit)
}
}
}
func printMatrix(randMatrix [][]int) {
//looping over 2D slice and extracting 1D slice to val
for _, val := range randMatrix {
fmt.Println(val)// printing each slice
}
}
We could modify the printMatrix function above, by looping over each integer and then formatting it well by using fmt.Printf(), but that would complicate things when we don't known the length of the limit...
I encountered weird behaviour in go code today: when I append elements to slice in loop and then try to create new slices based on the result of the loop, last append overrides slices from previous appends.
In this particular example it means that sliceFromLoop j,g and h slice's last element are not 100,101 and 102 respectively, but...always 102!
Second example - sliceFromLiteral behaves as expected.
package main
import "fmt"
func create(iterations int) []int {
a := make([]int, 0)
for i := 0; i < iterations; i++ {
a = append(a, i)
}
return a
}
func main() {
sliceFromLoop()
sliceFromLiteral()
}
func sliceFromLoop() {
fmt.Printf("** NOT working as expected: **\n\n")
i := create(11)
fmt.Println("initial slice: ", i)
j := append(i, 100)
g := append(i, 101)
h := append(i, 102)
fmt.Printf("i: %v\nj: %v\ng: %v\nh:%v\n", i, j, g, h)
}
func sliceFromLiteral() {
fmt.Printf("\n\n** working as expected: **\n")
i := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
fmt.Println("initial slice: ", i)
j := append(i, 100)
g := append(i, 101)
h := append(i, 102)
fmt.Printf("i: %v\nj: %v\ng: %v\nh:%v\n", i, j, g, h)
}
link to play.golang:
https://play.golang.org/p/INADVS3Ats
After some reading, digging and experimenting I found that this problem is originated in slices referencing the same underlaying array values and can be solved by copying slice to new one before appending anything, however it looks quite... hesitantly.
What's the idomatic way for creating many new slices based on old ones and not worrying about changing values of old slices?
Don't assign append to anything other than itself.
As you mention in the question, the confusion is due to the fact that append both changes the underlying array and returns a new slice (since the length might be changed). You'd imagine that it copies that backing array, but it doesn't, it just allocates a new slice object that points at it. Since i never changes, all those appends keep changing the value of backingArray[12] to a different number.
Contrast this to appending to an array, which allocates a new literal array every time.
So yes, you need to copy the slice before you can work on it.
func makeFromSlice(sl []int) []int {
result := make([]int, len(sl))
copy(result, sl)
return result
}
func main() {
i := make([]int, 0)
for ii:=0; ii<11; ii++ {
i = append(i, ii)
}
j := append(makeFromSlice(i), 100) // works fine
}
The slice literal behavior is explained because a new array is allocated if the append would exceed the cap of the backing array. This has nothing to do with slice literals and everything to do with the internals of how exceeding the cap works.
a := []int{1,2,3,4,5,6,7}
fmt.Printf("len(a) %d, cap(a) %d\n", a, len(a), cap(a))
// len(a) 7, cap(a) 7
b := make([]int, 0)
for i:=1; i<8, i++ {
b = append(b, i)
} // b := []int{1,2,3,4,5,6,7}
// len(b) 7, cap(b) 8
b = append(b, 1) // any number, just so it hits cap
i := append(b, 100)
j := append(b, 101)
k := append(b, 102) // these work as expected now
If you need a copy of a slice, there's no other way to do it other than, copying the slice. You should almost never assign the result of append to a variable other than the first argument of append. It leads to hard to find bugs, and will behave differently depending on whether the slice has the required capacity or not.
This isn't a commonly needed pattern, but as with all things of this nature if you need to repeate a few lines of code multiple times, then you can use a small helper function:
func copyAndAppend(i []int, vals ...int) []int {
j := make([]int, len(i), len(i)+len(vals))
copy(j, i)
return append(j, vals...)
}
https://play.golang.org/p/J99_xEbaWo
There is also a little bit simpler way to implement copyAndAppend function:
func copyAndAppend(source []string, items ...string) []string {
l := len(source)
return append(source[:l:l], items...)
}
Here we just make sure that source has no available capacity and so copying is forced.
I wanted to implement time based slots for holding data using golang slices. I managed to come up with a go program like this and it also works. But I have few questions regarding garbage collection and the general performance of this program. Does this program guarantee garbage collection of items once slice is equated to nil? And while shuffling slices, I hope this program does not do any deep copying.
type DataSlots struct {
slotDuration int //in milliseconds
slots [][]interface{}
totalDuration int //in milliseconds
}
func New(slotDur int, totalDur int) *DataSlots {
dat := &DataSlots{slotDuration: slotDur,
totalDuration: totalDur}
n := totalDur / slotDur
dat.slots = make([][]interface{}, n)
for i := 0; i < n; i++ {
dat.slots[i] = make([]interface{}, 0)
}
go dat.manageSlots()
return dat
}
func (self *DataSlots) addData(data interface{}) {
self.slots[0] = append(self.slots[0], data)
}
// This should be a go routine
func (self *DataSlots) manageSlots() {
n := self.totalDuration / self.slotDuration
for {
time.Sleep(time.Duration(self.slotDuration) * time.Millisecond)
for i := n - 1; i > 0; i-- {
self.slots[i] = self.slots[i-1]
}
self.slots[0] = nil
}
}
I removed critical section handling in this snippet to make it concise.
Once your slice is set too nil, any values contained in the slice are available for garbage collection, provided that the underlying array isn't shared with another slice.
Since there are no slice operations in your program, you never have multiple references to the same array, nor are you leaving data in any inaccessible portions of the underlying array.
What you need to be careful of, is when you're using slice operations:
a := []int{1, 2, 3, 4}
b := a[1:3]
a = nil
// the values 1 and 4 can't be collected, because they are
// still contained in b's underlying array
c := []int{1, 2, 3, 4}
c = append(c[1:2], 5)
// c is now []int{2, 5}, but again the values 1 and 4 are
// still in the underlying array. The 4 may be overwritten
// by a later append, but the 1 is inaccessible and won't
// be collected until the underlying array is copied.
While append does copy values when the capacity of the slice in insufficient, only the values contained in the slice are copied. There is no deep copy of any of the values.
What is the best way to remove items from a slice while ranging over it?
For example:
type MultiDataPoint []*DataPoint
func (m MultiDataPoint) Json() ([]byte, error) {
for i, d := range m {
err := d.clean()
if ( err != nil ) {
//Remove the DP from m
}
}
return json.Marshal(m)
}
As you have mentioned elsewhere, you can allocate new memory block and copy only valid elements to it. However, if you want to avoid the allocation, you can rewrite your slice in-place:
i := 0 // output index
for _, x := range s {
if isValid(x) {
// copy and increment index
s[i] = x
i++
}
}
// Prevent memory leak by erasing truncated values
// (not needed if values don't contain pointers, directly or indirectly)
for j := i; j < len(s); j++ {
s[j] = nil
}
s = s[:i]
Full example: http://play.golang.org/p/FNDFswPeDJ
Note this will leave old values after index i in the underlying array, so this will leak memory until the slice itself is garbage collected, if values are or contain pointers. You can solve this by setting all values to nil or the zero value from i until the end of the slice before truncating it.
I know its answered long time ago but i use something like this in other languages, but i don't know if it is the golang way.
Just iterate from back to front so you don't have to worry about indexes that are deleted. I am using the same example as Adam.
m = []int{3, 7, 2, 9, 4, 5}
for i := len(m)-1; i >= 0; i-- {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
}
}
There might be better ways, but here's an example that deletes the even values from a slice:
m := []int{1,2,3,4,5,6}
deleted := 0
for i := range m {
j := i - deleted
if (m[j] & 1) == 0 {
m = m[:j+copy(m[j:], m[j+1:])]
deleted++
}
}
Note that I don't get the element using the i, d := range m syntax, since d would end up getting set to the wrong elements once you start deleting from the slice.
Here is a more idiomatic Go way to remove elements from slices.
temp := s[:0]
for _, x := range s {
if isValid(x) {
temp = append(temp, x)
}
}
s = temp
Playground link: https://play.golang.org/p/OH5Ymsat7s9
Note: The example and playground links are based upon #tomasz's answer https://stackoverflow.com/a/20551116/12003457
One other option is to use a normal for loop using the length of the slice and subtract 1 from the index each time a value is removed. See the following example:
m := []int{3, 7, 2, 9, 4, 5}
for i := 0; i < len(m); i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
i-- // -1 as the slice just got shorter
}
}
I don't know if len() uses enough resources to make any difference but you could also run it just once and subtract from the length value too:
m := []int{3, 7, 2, 9, 4, 5}
for i, s := 0, len(m); i < s; i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
s--
i--
}
}
Something like:
m = append(m[:i], m[i+1:]...)
You don't even need to count backwards but you do need to check that you're at the end of the array where the suggested append() will fail. Here's an example of removing duplicate positive integers from a sorted list:
// Remove repeating numbers
numbers := []int{1, 2, 3, 3, 4, 5, 5}
log.Println(numbers)
for i, numbersCount, prevNum := 0, len(numbers), -1; i < numbersCount; numbersCount = len(numbers) {
if numbers[i] == prevNum {
if i == numbersCount-1 {
numbers = numbers[:i]
} else {
numbers = append(numbers[:i], numbers[i+1:]...)
}
continue
}
prevNum = numbers[i]
i++
}
log.Println(numbers)
Playground: https://play.golang.org/p/v93MgtCQsaN
I just implement a method which removes all nil elements in slice.
And I used it to solve a leetcode problems, it works perfectly.
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNil(lists *[]*ListNode) {
for i := 0; i < len(*lists); i++ {
if (*lists)[i] == nil {
*lists = append((*lists)[:i], (*lists)[i+1:]...)
i--
}
}
}
You can avoid memory leaks, as suggested in #tomasz's answer, controlling the capacity of the underlying array with a full slice expression. Look at the following function that remove duplicates from a slice of integers:
package main
import "fmt"
func removeDuplicates(a []int) []int {
for i, j := 0, 1; i < len(a) && j < len(a); i, j = i+1, j+1 {
if a[i] == a[j] {
copy(a[j:], a[j+1:])
// resize the capacity of the underlying array using the "full slice expression"
// a[low : high : max]
a = a[: len(a)-1 : len(a)-1]
i--
j--
}
}
return a
}
func main() {
a := []int{2, 3, 3, 3, 6, 9, 9}
fmt.Println(a)
a = removeDuplicates(a)
fmt.Println(a)
}
// [2 3 3 3 6 9 9]
// [2 3 6 9]
For reasons #tomasz has explained, there are issues with removing in place. That's why it is practice in golang not to do that, but to reconstruct the slice. So several answers go beyond the answer of #tomasz.
If elements should be unique, it's practice to use the keys of a map for this. I like to contribute an example of deletion by use of a map.
What's nice, the boolean values are available for a second purpose. In this example I calculate Set a minus Set b. As Golang doesn't have a real set, I make sure the output is unique. I use the boolean values as well for the algorithm.
The map gets close to O(n). I don't know the implementation. append() should be O(n). So the runtime is similar fast as deletion in place. Real deletion in place would cause a shifting of the upper end to clean up. If not done in batch, the runtime should be worse.
In this special case, I also use the map as a register, to avoid a nested loop over Set a and Set b to keep the runtime close to O(n).
type Set []int
func differenceOfSets(a, b Set) (difference Set) {
m := map[int]bool{}
for _, element := range a {
m[element] = true
}
for _, element := range b {
if _, registered := m[element]; registered {
m[element] = false
}
}
for element, present := range m {
if present {
difference = append(difference, element)
}
}
return difference
}
Try Sort and Binary search.
Example:
package main
import (
"fmt"
"sort"
)
func main() {
// Our slice.
s := []int{3, 7, 2, 9, 4, 5}
// 1. Iterate over it.
for i, v := range s {
func(i, v int) {}(i, v)
}
// 2. Sort it. (by whatever condition of yours)
sort.Slice(s, func(i, j int) bool {
return s[i] < s[j]
})
// 3. Cut it only once.
i := sort.Search(len(s), func(i int) bool { return s[i] >= 5 })
s = s[i:]
// That's it!
fmt.Println(s) // [5 7 9]
}
https://play.golang.org/p/LnF6o0yMJGT
http://play.golang.org/p/W70J4GU7nA
s := []int{5, 2, 6, 3, 1, 4}
sort.Reverse(sort.IntSlice(s))
fmt.Println(s)
// 5, 2, 6, 3, 1, 4
It is hard to understand what it means in func Reverse(data Interface) Interface .
How do I reverse an array? I do not need to sort.
Honestly this one is simple enough that I'd just write it out like this:
package main
import "fmt"
func main() {
s := []int{5, 2, 6, 3, 1, 4}
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
fmt.Println(s)
}
http://play.golang.org/p/vkJg_D1yUb
(The other answers do a good job of explaining sort.Interface and how to use it; so I won't repeat that.)
Normally, to sort an array of integers you wrap them in an IntSlice, which defines the methods Len, Less, and Swap. These methods are in turn used by sort.Sort. What sort.Reverse does is that it takes an existing type that defines Len, Less, and Swap, but it replaces the Less method with a new one that is always the inverse of the underlying Less:
type reverse struct {
// This embedded Interface permits Reverse to use the methods of
// another Interface implementation.
Interface
}
// Less returns the opposite of the embedded implementation's Less method.
func (r reverse) Less(i, j int) bool {
return r.Interface.Less(j, i)
}
// Reverse returns the reverse order for data.
func Reverse(data Interface) Interface {
return &reverse{data}
}
So when you write sort.Reverse(sort.IntSlice(s)), whats happening is that you're getting this new, 'modified' IntSlice that has it's Less method replaced. So if you call sort.Sort on it, which calls Less, it will get sorted in decreasing order.
I'm 2 years late, but just for fun and interest I'd like to contribute an "oddball" solution.
Assuming the task really is to reverse a list, then for raw performance bgp's solution is probably unbeatable. It gets the job done simply and effectively by swapping array items front to back, an operation that's efficient in the random-access structure of arrays and slices.
In Functional Programming languages, the idiomatic approach would often involve recursion. This looks a bit strange in Go and will have atrocious performance. That said, here's a recursive array reversal function (in a little test program):
package main
import (
"fmt"
)
func main() {
myInts := []int{ 8, 6, 7, 5, 3, 0, 9 }
fmt.Printf("Ints %v reversed: %v\n", myInts, reverseInts(myInts))
}
func reverseInts(input []int) []int {
if len(input) == 0 {
return input
}
return append(reverseInts(input[1:]), input[0])
}
Output:
Ints [8 6 7 5 3 0 9] reversed: [9 0 3 5 7 6 8]
Again, this is for fun and not production. Not only is it slow, but it will overflow the stack if the list is too large. I just tested, and it will reverse a list of 1 million ints but crashes on 10 million.
First of all, if you want to reverse the array, do like this,
for i, j := 0, len(a)-1; i < j; i, j = i+1, j-1 {
a[i], a[j] = a[j], a[i]
}
Then, look at the usage of Reverse in golang.org
package main
import (
"fmt"
"sort"
)
func main() {
s := []int{5, 2, 6, 3, 1, 4} // unsorted
sort.Sort(sort.Reverse(sort.IntSlice(s)))
fmt.Println(s)
}
// output
// [6 5 4 3 2 1]
And look at the description of Reverse and Sort
func Reverse(data Interface) Interface
func Sort(data Interface)
Sort sorts data. It makes one call to data.Len to determine n, and O(n*log(n)) calls to data.Less and data.Swap. The sort is not guaranteed to be stable.
So, as you know, Sort is not just a sort algorithm, you can view it as a factory, when you use Reverse it just return a reversed sort algorithm, Sort is just doing the sorting.
This is a more generic slice reverse function. It will panic if input is not a slice.
//panic if s is not a slice
func ReverseSlice(s interface{}) {
size := reflect.ValueOf(s).Len()
swap := reflect.Swapper(s)
for i, j := 0, size-1; i < j; i, j = i+1, j-1 {
swap(i, j)
}
}
If you want to reverse the array, you can just go through it in reverse order. Since there is no "reverse range" primitive in the language (at least not yet), you must do something like this (http://play.golang.org/p/AhvAfMjs_7):
s := []int{5, 2, 6, 3, 1, 4}
for i := len(s) - 1; i >= 0; i-- {
fmt.Print(s[i])
if i > 0 {
fmt.Print(", ")
}
}
fmt.Println()
Regarding whether it is hard to understand what sort.Reverse(data Interface) Interface does, I thought the same until I saw the source code from "http://golang.org/src/pkg/sort/sort.go".
It just makes the comparisons required for the sorting to be made "the other way around".
Here is a simple Go solution that uses an efficient (no extra memory) approach to reverse an array:
i := 0
j := len(nums) - 1
for i < j {
nums[i], nums[j] = nums[j], nums[i]
i++
j--
}
The idea is that reversing an array is equivalent to swapping each element with its mirror image across the center.
https://play.golang.org/p/kLFpom4LH0g
Here is another way to do it
func main() {
example := []int{1, 25, 3, 5, 4}
sort.SliceStable(example, func(i, j int) bool {
return true
})
fmt.Println(example)
}
https://play.golang.org/p/-tIzPX2Ds9z
func Reverse(data Interface) Interface
This means that it takes a sort.Interface and returns another sort.Interface -- it doesn't actually doing any sorting itself. For example, if you pass in sort.IntSlice (which is essentially a []int that can be passed to sort.Sort to sort it in ascending order) you'll get a new sort.Interface which sorts the ints in descending order instead.
By the way, if you click on the function name in the documentation, it links directly to the source for Reverse. As you can see, it just wraps the sort.Interface that you pass in, so the value returned from Reverse gets all the methods of the original sort.Interface. The only method that's different is the Less method which returns the opposite of the Less method on the embedded sort.Interface. See this part of the language spec for details on embedded fields.
From Golang wiki SliceTricks:
To replace the contents of a slice with the same elements but in
reverse order:
for i := len(a)/2-1; i >= 0; i-- {
opp := len(a)-1-i
a[i], a[opp] = a[opp], a[i]
}
The same thing, except with two indices:
for left, right := 0, len(a)-1; left < right; left, right = left+1, right-1 {
a[left], a[right] = a[right], a[left]
}
To reverse an array in place, iterate to its mid-point, and swap each element with its "mirror element":
func main() {
xs := []int{1, 2, 3, 4, 5, 6, 7, 8, 9}
itemCount := len(xs)
for i := 0; i < itemCount/2; i++ {
mirrorIdx := itemCount - i -1
xs[i], xs[mirrorIdx] = xs[mirrorIdx], xs[i]
}
fmt.Printf("xs: %v\n", xs)
}
https://play.golang.org/p/JeSApt80_k
Here is a method using append:
package main
import "fmt"
func main() {
a := []int{10, 20, 30, 40, 50}
for n := len(a) - 2; n >= 0; n-- {
a = append(a[:n], append(a[n + 1:], a[n])...)
}
fmt.Println(a)
}
Drawing of the steps:
10 20 30 40 50
10 20 30 50 40
10 20 50 40 30
10 50 40 30 20
50 40 30 20 10
This answer is mainly for those beginners who wish to write this code using only one variable in the for loop instead of using two variables (like i & j).
package main
import "fmt"
func main() {
array := []int{45, 17, 43, 67, 21, 4, 97, 44, 54, 98, 665}
fmt.Println("initial array:", array)
loop_iteration := len(array)
if len(array)%2 == 0 {
loop_iteration = (len(array) / 2) - 1
} else {
loop_iteration = int(len(array) / 2) //This will give the lower integer value of that float number.
}
for i := 0; i <= loop_iteration; i++ {
array[i], array[(len(array)-1)-i] = array[(len(array)-1)-i], array[i]
}
fmt.Println("reverse array:", array)
}
https://go.dev/play/p/bVp0x7v6Kbs
package main
import (
"fmt"
)
func main() {
arr := []int{1, 2, 3, 4, 5}
fmt.Println(reverseArray(arr))
}
func reverseArray(arr []int) []int {
reversed := make([]int, len(arr))
j := 0
for i := len(arr) - 1; i >= 0; i-- {
reversed[j] = arr[i]
j++
}
return reversed
}
Simple solution without involving math. Like this solution, this is inefficient as it does too much allocation and garbage collection. Good for non-critical code where clarity is more important than performance. Playground: https://go.dev/play/p/dQGwrc0Q9ZA
arr := []int{1, 3, 4, 5, 6}
var rev []int
for _, n := range arr {
rev = append([]int{n}, rev...)
}
fmt.Println(arr)
fmt.Println(rev)
Its very simple if you want to print reverse array
Use Index from length doing i--
ex.
a := []int{5, 4, 12, 7, 15, 9}
for i := 0; i <= len(a)-1; i++ {
fmt.Println(a[len(a)-(i+1)])
}
https://go.dev/play/p/bmyFh7-_VCZ
Here is my solution.
package main
import (
"fmt"
)
func main() {
var numbers = [10]int {1,2,3,4,5,6,7,8,9,10}
var reverseNumbers [10]int
j:=0
for i:=len(numbers)-1; i>=0 ; i-- {
reverseNumbers[j]=numbers[i]
j++
}
fmt.Println(reverseNumbers)
}
Here is my solution to reversing an array:
func reverse_array(array []string) []string {
lenx := len(array) // lenx holds the original array length
reversed_array := make([]string, lenx) // creates a slice that refer to a new array of length lenx
for i := 0; i < lenx; i++ {
j := lenx - (i + 1) // j initially holds (lenx - 1) and decreases to 0 while i initially holds 0 and increase to (lenx - 1)
reversed_array[i] = array[j]
}
return reversed_array
}
You can try this solution on the go playground the go playground
package main
import "fmt"
func main() {
array := []string{"a", "b", "c", "d"}
fmt.Println(reverse_array(array)) // prints [d c b a]
}
Do not reverse it, leave it as now and then just iterate it backwards.