Array Reversal Techniques in DSA C - Execution Trace

Concept Flow - Array Reversal Techniques

Start with array

↓

Set left = 0, right = length-1

↓

Check if left < right?

No→Done

Yes↓

Swap elements at left and right

↓

Increment left, Decrement right

↩Back to Check

The process swaps elements from the ends moving inward until the middle is reached.

Execution Sample

DSA C

int arr[] = {1, 2, 3, 4, 5};
int left = 0, right = 4;
while (left < right) {
  int temp = arr[left];
  arr[left] = arr[right];
  arr[right] = temp;
  left++;
  right--;
}

This code reverses the array by swapping elements from the start and end moving towards the center.

Execution Table

Step	Operation	left	right	Array State	Pointer Changes
1	Initialize left and right	0	4	[1, 2, 3, 4, 5]	left=0, right=4
2	Check left < right	0	4	[1, 2, 3, 4, 5]	0 < 4 is True
3	Swap arr[left] and arr[right]	0	4	[5, 2, 3, 4, 1]	Swapped arr[0] and arr[4]
4	Increment left, Decrement right	1	3	[5, 2, 3, 4, 1]	left=1, right=3
5	Check left < right	1	3	[5, 2, 3, 4, 1]	1 < 3 is True
6	Swap arr[left] and arr[right]	1	3	[5, 4, 3, 2, 1]	Swapped arr[1] and arr[3]
7	Increment left, Decrement right	2	2	[5, 4, 3, 2, 1]	left=2, right=2
8	Check left < right	2	2	[5, 4, 3, 2, 1]	2 < 2 is False
9	Done	2	2	[5, 4, 3, 2, 1]	Reversal complete

💡 left is not less than right (2 < 2 is False), so loop ends

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	After Step 7	Final
left	0	0	1	1	2	2
right	4	4	3	3	2	2
arr	[1, 2, 3, 4, 5]	[5, 2, 3, 4, 1]	[5, 2, 3, 4, 1]	[5, 4, 3, 2, 1]	[5, 4, 3, 2, 1]	[5, 4, 3, 2, 1]

Key Moments - 3 Insights

Why do we stop when left is not less than right?

Because when left equals or passes right, all pairs have been swapped. See execution_table rows 8 and 9 where 2 < 2 is False, so reversal is complete.

Why do we swap elements at left and right instead of just reversing by copying?

Swapping in place saves memory by not needing extra space. The execution_table shows how elements at positions left and right are exchanged step-by-step.

What happens if the array has an odd number of elements?

The middle element stays in place because left and right meet at the center. In the example, arr[2] = 3 remains unchanged as shown in the array state after step 7.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the array state?

A [5, 4, 3, 2, 1]

B [5, 2, 3, 4, 1]

C [1, 2, 3, 4, 5]

D [1, 4, 3, 2, 5]

💡 Hint

Check the 'Array State' column at step 6 in the execution_table.

At which step does the condition left < right become false?

A Step 7

B Step 8

C Step 5

D Step 9

💡 Hint

Look at the 'Pointer Changes' and 'Operation' columns in execution_table rows 8 and 9.

If the array length was 6 instead of 5, how would the final left and right values differ?

A left=2, right=3

B left=2, right=2

C left=3, right=2

D left=3, right=3

💡 Hint

Consider how left increments and right decrements until left >= right for even length arrays.

Concept Snapshot

Array Reversal Technique:
- Use two pointers: left at start, right at end
- While left < right:
  - Swap arr[left] and arr[right]
  - Increment left, decrement right
- Stops when pointers meet or cross
- Works in-place without extra memory

Full Transcript

This visualization shows how to reverse an array by swapping elements from the ends moving inward. We start with two pointers, left at index 0 and right at the last index. While left is less than right, we swap the elements at these positions, then move left forward and right backward. This continues until left is no longer less than right, meaning the array is reversed. The example array [1, 2, 3, 4, 5] becomes [5, 4, 3, 2, 1] after the process. Key points include stopping when pointers meet or cross, and that the middle element in odd-length arrays stays in place. This method reverses the array in-place without extra memory.