Remove Duplicates from Sorted Array Two Pointer in DSA Python - Execution Trace

Concept Flow - Remove Duplicates from Sorted Array Two Pointer

Start with two pointers

↓

Pointer1 at index 0

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Compare values at Pointer1 and Pointer2

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Repeat until Pointer2 reaches end

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Return length = Pointer1 + 1

Use two pointers to scan the array. One pointer tracks unique elements, the other scans ahead to find new unique values to copy forward.

Execution Sample

DSA Python

nums = [1,1,2,2,3]
left = 0
for right in range(1, len(nums)):
    if nums[right] != nums[left]:
        left += 1
        nums[left] = nums[right]
print(left + 1)

This code removes duplicates in-place from a sorted array and returns the new length of unique elements.

Execution Table

Step	Operation	Pointer Left	Pointer Right	Array State	Action
0	Initialize pointers	0	1	[1,1,2,2,3]	Start with left=0, right=1
1	Compare nums[left] and nums[right]	0	1	[1,1,2,2,3]	1 == 1, duplicate found, move right
2	Move right pointer	0	2	[1,1,2,2,3]	right moves to 2
3	Compare nums[left] and nums[right]	0	2	[1,1,2,2,3]	1 != 2, unique found
4	Move left pointer and copy	1	2	[1,2,2,2,3]	left moves to 1, nums[1] = 2
5	Move right pointer	1	3	[1,2,2,2,3]	right moves to 3
6	Compare nums[left] and nums[right]	1	3	[1,2,2,2,3]	2 == 2, duplicate found, move right
7	Move right pointer	1	4	[1,2,2,2,3]	right moves to 4
8	Compare nums[left] and nums[right]	1	4	[1,2,2,2,3]	2 != 3, unique found
9	Move left pointer and copy	2	4	[1,2,3,2,3]	left moves to 2, nums[2] = 3
10	Right pointer reached end	2	4	[1,2,3,2,3]	End of array reached
11	Return length	2	4	[1,2,3,2,3]	Length = left + 1 = 3

💡 Right pointer reached end of array, all unique elements moved to front

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 7	After Step 9	Final
left	0	0	1	1	1	2	2
right	1	1	2	3	4	4	4
nums	[1,1,2,2,3]	[1,1,2,2,3]	[1,2,2,2,3]	[1,2,2,2,3]	[1,2,2,2,3]	[1,2,3,2,3]	[1,2,3,2,3]

Key Moments - 3 Insights

Why do we move the left pointer only when we find a new unique element?

Because left pointer marks the position of last unique element. We only advance it to store the next unique value, as shown in steps 4 and 9 in the execution_table.

Why don't we move the left pointer when duplicates are found?

When duplicates are found (steps 1 and 6), we only move the right pointer to skip them. Left pointer stays to overwrite duplicates later with new unique values.

Why is the returned length left + 1?

Because left is zero-based index of last unique element, so length is left + 1 as shown in step 11.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the array state?

A [1,2,3,2,3]

B [1,2,2,2,3]

C [1,1,2,2,3]

D [1,1,3,2,3]

💡 Hint

Check the 'Array State' column at step 4 in execution_table

At which step does the left pointer move for the second time?

A Step 4

B Step 6

C Step 9

D Step 2

💡 Hint

Look at the 'Pointer Left' column in execution_table and find when it changes the second time

If the input array was [1,1,1,1], what would be the returned length?

A 1

B 4

C 0

D 2

💡 Hint

All elements are duplicates except the first, so left pointer stays at 0, length = left + 1

Concept Snapshot

Remove duplicates from sorted array using two pointers:
- left pointer tracks last unique element
- right pointer scans ahead
- if nums[right] != nums[left], move left and copy nums[right]
- continue until right reaches end
- return left + 1 as new length

Full Transcript

This visualization shows how to remove duplicates from a sorted array using two pointers. We start with left pointer at index 0 and right pointer at index 1. We compare values at these pointers. If they are the same, right pointer moves forward to skip duplicates. If different, left pointer moves forward and copies the unique value from right pointer. This process continues until right pointer reaches the end of the array. The final length of unique elements is left pointer index plus one. The array is modified in place to have unique elements at the front.