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.

Practice

1. What is the main idea behind the Bubble Sort algorithm?

easy

A. Repeatedly swap adjacent elements if they are in the wrong order

B. Divide the list into halves and sort each half separately

C. Select the smallest element and move it to the beginning

D. Insert each element into its correct position in a sorted sublist

Solution

1. Step 1: Understand Bubble Sort mechanism

   Bubble Sort compares adjacent elements and swaps them if they are out of order.

2. Step 2: Compare with other sorting methods

   Other options describe different algorithms: divide and conquer, selection sort, insertion sort.

3. Final Answer:

   Repeatedly swap adjacent elements if they are in the wrong order -> Option A

4. Quick Check:

   Bubble Sort = swapping adjacent pairs [OK]

Hint: Bubble Sort swaps neighbors repeatedly until sorted [OK]

Common Mistakes:

• Confusing Bubble Sort with Selection Sort
• Thinking it divides the list like Merge Sort
• Assuming it inserts elements like Insertion Sort

2. Which of the following is the correct way to write the outer loop for Bubble Sort in Go to sort an array arr of length n?

easy

A. for i := 0; i < n-1; i++ {}

B. for i := 0; i < n; i++ {}

C. for i := 1; i <= n; i++ {}

D. for i := 1; i < n-1; i++ {}

Solution

1. Step 1: Determine loop range for passes

   Bubble Sort requires n-1 passes to ensure sorting, so loop runs from 0 to n-2 (i < n-1).

2. Step 2: Check options for correct syntax and range

   for i := 0; i < n-1; i++ {} correctly uses i := 0; i < n-1; i++. Others either run too many or incorrect iterations.

3. Final Answer:

   for i := 0; i < n-1; i++ {} -> Option A

4. Quick Check:

   Outer loop runs n-1 times [OK]

Hint: Outer loop runs n-1 times in Bubble Sort [OK]

Common Mistakes:

• Using i <= n causing out-of-range errors
• Starting loop at 1 instead of 0
• Looping n times instead of n-1

3. What will be the output of the following Go code snippet after applying one full pass of Bubble Sort on the array?

arr := []int{5, 3, 8, 4, 2}
n := len(arr)
for j := 0; j < n-1; j++ {
    if arr[j] > arr[j+1] {
        arr[j], arr[j+1] = arr[j+1], arr[j]
    }
}
fmt.Println(arr)

medium

A. [2 3 4 5 8]

B. [3 5 8 4 2]

C. [3 5 4 2 8]

D. [5 3 8 4 2]

Solution

1. Step 1: Perform one pass of Bubble Sort

   Compare and swap adjacent elements: - Compare 5 & 3: swap -> [3 5 8 4 2] - Compare 5 & 8: no swap - Compare 8 & 4: swap -> [3 5 4 8 2] - Compare 8 & 2: swap -> [3 5 4 2 8]

2. Step 2: Final array after one pass

   After one pass, largest element 8 moves to the end. Array is [3 5 4 2 8].

3. Final Answer:

   [3 5 4 2 8] -> Option C

4. Quick Check:

   One pass moves largest to end [OK]

Hint: One pass pushes largest element to the end [OK]

Common Mistakes:

• Assuming full sort after one pass
• Not swapping correctly
• Confusing indices in swaps

4. Identify the error in the following Bubble Sort code snippet in Go:

arr := []int{4, 2, 7, 1}
n := len(arr)
for i := 0; i < n; i++ {
    for j := 0; j < n; j++ {
        if arr[j] > arr[j+1] {
            arr[j], arr[j+1] = arr[j+1], arr[j]
        }
    }
}
fmt.Println(arr)

medium

A. Outer loop should start from 1 instead of 0

B. Inner loop runs out of range causing runtime error

C. Swap condition should be arr[j] < arr[j+1]

D. Array length n is incorrect

Solution

1. Step 1: Analyze inner loop range

   Inner loop runs j from 0 to n-1, but accesses arr[j+1]. When j = n-1, arr[j+1] is out of range causing runtime error.

2. Step 2: Correct inner loop range

   Inner loop should run from 0 to n-2 (j < n-1) to avoid out-of-range access.

3. Final Answer:

   Inner loop runs out of range causing runtime error -> Option B

4. Quick Check:

   Index out of range error = inner loop limit [OK]

Hint: Inner loop must stop before last index to avoid out-of-range [OK]

Common Mistakes:

• Not adjusting inner loop limit
• Swapping with wrong comparison
• Misunderstanding loop boundaries

5. Given the array arr := []int{3, 1, 4, 5, 2}, how many total swaps will Bubble Sort perform to fully sort the array in ascending order?

hard

A. 5

B. 7

C. 6

D. 4

Solution

1. Step 1: Trace swaps in each pass

   Initial array: [3,1,4,5,2] Pass 1: - 3 > 1 swap -> [1,3,4,5,2] (1 swap) - 3 > 4 no swap - 4 > 5 no swap - 5 > 2 swap -> [1,3,4,2,5] (2 swaps total) Pass 2: - 1 > 3 no swap - 3 > 4 no swap - 4 > 2 swap -> [1,3,2,4,5] (3 swaps total) Pass 3: - 1 > 3 no swap - 3 > 2 swap -> [1,2,3,4,5] (4 swaps total) Pass 4: - 1 > 2 no swap - 2 > 3 no swap No swaps, sorting done.

2. Step 2: Count total swaps

   Pass 1: 2 swaps Pass 2: 1 swap Pass 3: 1 swap Total = 4 swaps Recheck carefully: Pass 1: swaps at (3,1) and (5,2) = 2 Pass 2: swap at (4,2) = 1 Pass 3: swap at (3,2) = 1 Total swaps = 4

3. Step 3: Verify if more swaps needed

   Sorting completes after pass 4 with no swaps. Total swaps = 4

4. Final Answer:

   4 -> Option D

5. Quick Check:

   Count swaps carefully per pass [OK]

Hint: Count swaps each pass until no swaps remain [OK]

Common Mistakes:

• Counting comparisons instead of swaps
• Missing swaps in later passes
• Assuming swaps equal to array length