# Write a Golang program to sort a binary array in linear time

There are two methods in which we can solve this problem. Let’s check the first method.

Method 1

## Examples

• Input Array = [1, 0, 1, 0, 1, 0, 0, 1] => [0, 0, 0, 0, 1, 1, 1, 1]

## Approach to solve this problem

Step 1: Define a method that accepts an array.

Step 2: Count number of 0.

Step 3: Store 0 till count becomes 0 and store 1 at the remaining indexes.

Step 4: At the end, return the array.

## Program

Live Demo

package main
import "fmt"
func binarySort(arr []int) []int{
count := 0
for i:=0; i<len(arr); i++{
if arr[i]==0{
count++
}
}
for j:=0; j<len(arr); j++{
if j<count{
arr[j] = 0
} else {
arr[j] = 1
}
}
return arr
}

func main(){
fmt.Println(binarySort([]int{1, 0, 1, 0, 1, 0, 0, 1}))
fmt.Println(binarySort([]int{1, 1, 1, 1, 1, 1, 1, 1}))
fmt.Println(binarySort([]int{0, 0, 0, 0, 0, 0, 0, 0}))
}

## Output

[0 0 0 0 1 1 1 1]
[1 1 1 1 1 1 1 1]
[0 0 0 0 0 0 0 0]

Method 2

Now, let’s check the second method.

## Approach to solve this problem

• Step 1: Define a method that accepts an array.
• Step 2: Declare the pivot element and its index j.
• Step 3: Iterate the given array. If the element is less than pivot, then swap and increase the pivot’s index.
• Step 4: At the end, return the array.

## Program

Live Demo

package main
import "fmt"
func binarySort(arr []int) []int{
pivot := 1
index := 0
for i:=0; i<len(arr); i++ {
if arr[i]<pivot{
arr[i], arr[index] = arr[index], arr[i]
index++
}
}
return arr
}

func main(){
fmt.Println(binarySort([]int{1, 0, 1, 0, 1, 0, 0, 1}))
fmt.Println(binarySort([]int{1, 1, 1, 1, 1, 1, 1, 1}))
fmt.Println(binarySort([]int{0, 0, 0, 0, 0, 0, 0, 0}))
}

## Output

[0 0 0 0 1 1 1 1]
[1 1 1 1 1 1 1 1]
[0 0 0 0 0 0 0 0]

Updated on: 04-Feb-2021

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