In bubble sort:

• In each iteration i (starting with 0) , we start from first element and repeatedly swap adjacent elements if in wrong order up till length (len-i) where len is the length of the array.

• At the end of the iteration, either the largest or smallest element (depending upon order is ascending or descending) is at position (len-i)

Time Complexity

• O(n*n)

Space Complexity

• Space Complexity of bubble sort is O(1)

Implementation:

package main

import "fmt"

func main() {
    sample := []int{3, 4, 5, 2, 1}
    bubbleSort(sample)
    sample = []int{3, 4, 5, 2, 1, 7, 8, -1, -3}
    bubbleSort(sample)
}

func bubbleSort(arr []int) {
    len := len(arr)
    for i := 0; i < len-1; i++ {
        for j := 0; j < len-i-1; j++ {
            if arr[j] > arr[j+1] {
                arr[j], arr[j+1] = arr[j+1], arr[j]
            }
        }
    }
    fmt.Println("\nAfter Bubble Sorting")
    for _, val := range arr {
        fmt.Println(val)
    }
}

Output:

After Bubble Sorting
1
2
3
4
5

After Bubble Sorting
-3
-1
1
2
3
4
5
7
8

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