Array Reversal Techniques in DSA Python - Execution Trace

Concept Flow - Array Reversal Techniques

Start with array

↓

Set left = 0, right = len-1

↓

Check if left < right?

No→Done

Yes↓

Swap elements at left and right

↓

Increment left, Decrement right

↩Back to check left < right

We start with two pointers at the ends of the array and swap elements moving inward until they meet.

Execution Sample

DSA Python

arr = [1, 2, 3, 4, 5]
left, right = 0, len(arr)-1
while left < right:
    arr[left], arr[right] = arr[right], arr[left]
    left += 1
    right -= 1
print(arr)

This code reverses the array by swapping elements from the ends moving towards the center.

Execution Table

Step	Operation	left	right	Array State	Pointer Changes
1	Initialize pointers	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]	Swap 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]	Swap 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	End loop	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 4	After Step 7	After Step 9 (Final)
left	0	1	2	2
right	4	3	2	2
arr	[1, 2, 3, 4, 5]	[5, 2, 3, 4, 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 meets or passes right, all elements have been swapped. See execution_table rows 8 and 9 where 2 < 2 is False, so we stop.

Why do we swap arr[left] and arr[right] instead of just moving pointers?

Swapping arr[left] and arr[right] reverses the elements at the ends. Without swapping, the array stays the same. See rows 3 and 6 where swaps change the array.

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. See row 7 where left and right both become 2, the middle index.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the array state after the swap?

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

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

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

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

💡 Hint

Check the 'Array State' column at Step 3 in execution_table.

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

A Step 6

B Step 4

C Step 8

D Step 2

💡 Hint

Look at the 'Pointer Changes' and 'Operation' columns in execution_table for the condition check.

If the array length was even, how would the final left and right pointers compare?

A left would be equal to right

B left would be greater than right

C left would be less than right

D left and right would not change

💡 Hint

Refer to variable_tracker and think about how pointers move inward during swaps.

Concept Snapshot

Array Reversal Technique:
- Use two pointers: left at start, right at end
- While left < right:
  - Swap arr[left] and arr[right]
  - Move left forward, right backward
- Stop when pointers meet or cross
- Works for arrays of any length

Full Transcript

This visual execution shows how to reverse an array by swapping elements from the ends moving inward. We start with two pointers, left at 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. When left meets or passes right, the array is reversed. The example uses the array [1, 2, 3, 4, 5] and shows each swap and pointer movement step by step, ending with the reversed array [5, 4, 3, 2, 1].