Bubble Sort Algorithm in DSA Go - Execution Trace

Concept Flow - Bubble Sort Algorithm

Start with array

↓

Compare adjacent elements

↓

If left > right?

No→Move to next pair

|Yes↓

Swap elements

↓

Reach end of array?

No→Repeat compare-swap

|Yes↓

Reduce array size by 1

↓

Array sorted?

No→Repeat passes

|Yes↓

Done

Bubble Sort repeatedly compares and swaps adjacent elements, pushing the largest unsorted element to the end each pass, until the array is sorted.

Execution Sample

DSA Go

arr := []int{5, 3, 8, 4}
for i := 0; i < len(arr)-1; i++ {
  for j := 0; j < len(arr)-1-i; j++ {
    if arr[j] > arr[j+1] {
      arr[j], arr[j+1] = arr[j+1], arr[j]
    }
  }
}

This code sorts the array using Bubble Sort by repeatedly swapping adjacent elements if they are in the wrong order.

Execution Table

Step	Operation	Comparison	Swap?	Array State
1.1	Compare arr[0] and arr[1]	5 > 3	Yes	[5, 3, 8, 4] → [3, 5, 8, 4]
1.2	Compare arr[1] and arr[2]	5 > 8	No	[3, 5, 8, 4]
1.3	Compare arr[2] and arr[3]	8 > 4	Yes	[3, 5, 8, 4] → [3, 5, 4, 8]
End of Pass 1	Largest element 8 placed at end	-	-	[3, 5, 4, 8]
2.1	Compare arr[0] and arr[1]	3 > 5	No	[3, 5, 4, 8]
2.2	Compare arr[1] and arr[2]	5 > 4	Yes	[3, 5, 4, 8] → [3, 4, 5, 8]
End of Pass 2	Second largest element 5 placed	-	-	[3, 4, 5, 8]
3.1	Compare arr[0] and arr[1]	3 > 4	No	[3, 4, 5, 8]
End of Pass 3	Array sorted	-	-	[3, 4, 5, 8]
Exit	No swaps needed, sorting complete	-	-	[3, 4, 5, 8]

💡 No swaps in pass 3, array is sorted

Variable Tracker

Variable	Start	After Pass 1	After Pass 2	After Pass 3	Final
arr	[5, 3, 8, 4]	[3, 5, 4, 8]	[3, 4, 5, 8]	[3, 4, 5, 8]	[3, 4, 5, 8]
i (pass index)	0	1	2	3	3
j (compare index)	-	0 to 2	0 to 1	0	-

Key Moments - 3 Insights

Why do we reduce the inner loop length by 'i' each pass?

Because after each pass, the largest element is at the end, so we don't need to compare it again. See execution_table rows 'End of Pass 1' and 'End of Pass 2' where largest elements are placed.

What happens if no swaps occur during a pass?

It means the array is already sorted and we can stop early. This is shown in execution_table row 'Exit' where no swaps happen in pass 3.

Why do we compare arr[j] and arr[j+1] instead of other pairs?

Bubble Sort compares adjacent pairs to push larger elements rightwards step by step. This is shown in every 'Compare' operation in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 1.3, what is the array state after the swap?

A [3, 5, 4, 8]

B [3, 5, 8, 4]

C [5, 3, 8, 4]

D [3, 4, 5, 8]

💡 Hint

Check the 'Array State' column in execution_table row 1.3

At which pass does the array become fully sorted according to the execution_table?

A Pass 1

B Pass 2

C Pass 3

D Pass 4

💡 Hint

Look at the 'End of Pass 3' row showing the sorted array

If the initial array was already sorted, how would the execution_table change?

A No swaps would occur in the first pass

B Swaps would occur in every pass

C The array state would change after each comparison

D The algorithm would run more passes

💡 Hint

Refer to the 'Exit' row where no swaps mean sorting is complete early

Concept Snapshot

Bubble Sort Algorithm:
- Repeatedly compare adjacent elements
- Swap if left element is greater
- Largest unsorted element moves to end each pass
- Reduce inner loop length each pass
- Stop early if no swaps in a pass

Full Transcript

Bubble Sort works by comparing pairs of adjacent elements in an array. If the left element is bigger, they swap places. This pushes the largest element to the end of the array after each pass. Each pass reduces the number of elements to check because the end elements are sorted. If a pass completes with no swaps, the array is sorted and the algorithm stops early. The execution table shows each comparison and swap step, tracking the array state until fully sorted.