Here, we will learn about algorithm and program of shifting all the array’s zero elements to the end of the array in go golang. There is a given array having random numbers. We have to move all the zero’s to the end of array. Time complexity is o(n) and space complexity is o(1).
Example:
Input: array = [1, 2, 0, 5, 0, 0, 3, 5]
Output: array = [1, 2, 5, 3, 5, 0, 0 ,0]
Input: array = [1, 2, 0, 0, 0, 0, 0, 2, 5]
Output: array = [1, 2, 2, 5, 0, 0, 0, 0, 0]
Algorithm:
1: Take two variable lets say index and count and initialise to 0 (zero).
2: Traverse the array using these two variables.
3: When arr[index] will not be zero, copy the value of arr[index] to arr[count] and increment both the variables by 1.
4: If arr[index] is zero then increment only index variable by 1.
5: At last, Fill the remaining array count index to zero.
For the more depth, Please follow the given below program.
Program:
package main
import (
"fmt"
)
func main() {
var no_of_array_ele, count int;
fmt.Println("Enter number of elements in array: ")
// Number of array elements
fmt.Scan(&no_of_array_ele)
// Create an array
arr := make([]int, no_of_array_ele)
// Insert the elements into array
for index := 0; index < no_of_array_ele; index++ {
fmt.Printf("\nEnter the %d element: ", index+1)
fmt.Scan(&arr[index]);
}
count = 0
for index := 0; index < no_of_array_ele; index++ {
if (arr[index] != 0) {
arr[count] = arr[index];
count = count + 1
}
}
// Fill the remaining array count index to zero
for count != no_of_array_ele {
arr[count] = 0;
count = count + 1
}
fmt.Println("Array elements after shifting all zero to ends: \n")
for index := 0; index < no_of_array_ele; index++ {
fmt.Printf("%d ", arr[index])
}
}
Enter number of elements in array: 5
Enter the 1 element: 1
Enter the 2 element: 0
Enter the 3 element: 2
Enter the 4 element: 0
Enter the 5 element: 4
Array elements before shifting all zero to ends:
1 0 2 0 4
Array elements after shifting all zero to ends:
1 2 4 0 0
For more details about golang, Please follow given below link.