Array Deletion at Middle Index in DSA Python - Execution Trace

Concept Flow - Array Deletion at Middle Index

Start with array

↓

Identify index to delete

↓

Shift elements left from index+1 to end

↓

Reduce array size by 1

↓

Resulting array without deleted element

We start with an array, find the index to delete, shift all elements after it one step left, then reduce the array size by one.

Execution Sample

DSA Python

arr = [10, 20, 30, 40, 50]
index = 2
for i in range(index, len(arr)-1):
    arr[i] = arr[i+1]
arr.pop()

Deletes element at index 2 by shifting elements left and removing last duplicate.

Execution Table

Step	Operation	Index i	Array State	Pointer Changes	Visual State
1	Start	-	[10, 20, 30, 40, 50]	head at 0	10 -> 20 -> 30 -> 40 -> 50 -> null
2	Delete at index	2	[10, 20, 30, 40, 50]	target index = 2	10 -> 20 -> 30 -> 40 -> 50 -> null
3	Shift element	2	[10, 20, 40, 40, 50]	arr[2] = arr[3]	10 -> 20 -> 40 -> 40 -> 50 -> null
4	Shift element	3	[10, 20, 40, 50, 50]	arr[3] = arr[4]	10 -> 20 -> 40 -> 50 -> 50 -> null
5	Pop last element	-	[10, 20, 40, 50]	size reduced by 1	10 -> 20 -> 40 -> 50 -> null
6	End	-	[10, 20, 40, 50]	-	10 -> 20 -> 40 -> 50 -> null

💡 All elements shifted left from index 2, last element removed to reduce size.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
arr	[10, 20, 30, 40, 50]	[10, 20, 40, 40, 50]	[10, 20, 40, 50, 50]	[10, 20, 40, 50]	[10, 20, 40, 50]
index	2	2	2	2	2
i	-	2	3	-	-

Key Moments - 3 Insights

Why do we shift elements left instead of just removing the element at the index?

Because arrays have fixed positions, removing an element leaves a gap. Shifting left fills that gap, keeping elements contiguous as shown in steps 3 and 4 of the execution_table.

Why do we pop the last element after shifting?

After shifting, the last element is duplicated. Popping removes this duplicate and reduces the array size, as shown in step 5.

What happens if we delete the last element (index = last)?

No shifting is needed; we just pop the last element. This is a special case not shown here but follows the same pop step.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after shifting element at index 2?

A [10, 20, 30, 40, 50]

B [10, 20, 40, 40, 50]

C [10, 20, 40, 50, 50]

D [10, 20, 40, 50]

💡 Hint

Check row with Step 3 in execution_table under 'Array State'

At which step does the array size reduce?

A Step 5

B Step 4

C Step 3

D Step 6

💡 Hint

Look for 'Pop last element' operation in execution_table

If we delete element at index 4 (last element), how would the execution_table change?

A Shift elements twice then pop

B Shift elements once then pop

C No shifting steps, only pop last element

D No pop needed

💡 Hint

Deleting last element means no elements after it to shift, so only pop is needed

Concept Snapshot

Array Deletion at Middle Index:
- Identify index to delete
- Shift all elements after index one step left
- Remove last element to reduce size
- Keeps array contiguous without gaps
- Works in O(n) time for shifting

Full Transcript

This visualization shows how to delete an element from the middle of an array. We start with the array and the index to delete. Then, we shift each element after that index one position to the left to fill the gap. Finally, we remove the last element since it is now duplicated. This keeps the array contiguous and reduces its size by one. The execution table tracks each step, showing how the array changes and how pointers move. Key moments clarify why shifting and popping are necessary. The quiz tests understanding of these steps.