Why Remove Duplicates from Sorted Array Two Pointer in DSA Python?

The Big Idea

What if you could clean a long list of duplicates in just one quick pass?

The Scenario

Imagine you have a long list of names sorted alphabetically, but some names appear more than once. You want to make a clean list with each name only once.

Doing this by checking each name one by one and removing duplicates manually is tiring and slow.

The Problem

Manually scanning the list and deleting duplicates means you have to move many names around each time you find a repeat.

This wastes time and can cause mistakes like skipping names or deleting the wrong one.

The Solution

The two-pointer method lets you scan the list just once, keeping track of unique names as you go.

One pointer moves through the list, the other marks where to place the next unique name.

This way, duplicates are skipped automatically, and the list is cleaned efficiently.

Before vs After

✗ Before

for i in range(len(names) - 1):
    if names[i] == names[i+1]:
        names.pop(i+1)

✓ After

unique_index = 0
for current_index in range(1, len(names)):
    if names[current_index] != names[unique_index]:
        unique_index += 1
        names[unique_index] = names[current_index]

What It Enables

This method makes it easy to clean sorted lists quickly, saving time and avoiding errors.

Real Life Example

When you get a sorted list of email addresses from a newsletter signup, you can quickly remove duplicates before sending emails.

Key Takeaways

Manual removal is slow and error-prone.

Two-pointer method scans once and keeps unique items.

Efficient and simple for sorted lists.

Practice

1. What is the main purpose of using two pointers in the Remove Duplicates from Sorted Array problem?

easy

A. To create a new array with unique elements

B. To sort the array before removing duplicates

C. To count the total number of duplicates in the array

D. To efficiently track unique elements and overwrite duplicates in-place

Solution

1. Step 1: Understand the two pointer roles

   One pointer tracks the position to place the next unique element, the other scans the array.

2. Step 2: Explain in-place overwrite

   When a new unique element is found, it is placed at the first pointer's position, avoiding extra space.

3. Final Answer:

   To efficiently track unique elements and overwrite duplicates in-place -> Option D

4. Quick Check:

   Two pointers help remove duplicates without extra space [OK]

Hint: Two pointers help overwrite duplicates without extra space [OK]

Common Mistakes:

• Thinking two pointers sort the array
• Assuming a new array is created
• Confusing counting duplicates with removing them

2. Which of the following is the correct syntax to initialize two pointers i and j for the two-pointer approach in Python?

easy

A. i = 1; j = 0

B. i = j = 0

C. i, j = 0, 1

D. i = 0; j = 0

Solution

1. Step 1: Identify pointer roles

   Pointer i starts at 0 (position for unique elements), j starts at 1 (scanning pointer).

2. Step 2: Match initialization

   Correct initialization is i, j = 0, 1 to start comparing adjacent elements.

3. Final Answer:

   i, j = 0, 1 -> Option C

4. Quick Check:

   Start pointers at 0 and 1 for two-pointer scan [OK]

Hint: Start pointers at 0 and 1 to compare adjacent elements [OK]

Common Mistakes:

• Initializing both pointers at 0
• Swapping pointer start positions
• Using incorrect syntax for multiple assignments

3. What will be the output of the following code snippet?

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

medium

A. [1, 1, 2, 3]

B. [1, 2, 3]

C. [1, 2, 2, 3]

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

Solution

1. Step 1: Trace the loop and pointer updates

   Initially, i=0, j scans from 1 to 4. When nums[j] != nums[i], i increments and nums[i] is updated.

2. Step 2: Track array changes

   After processing, nums becomes [1, 2, 3, 2, 3], but only first i+1=3 elements are unique.

3. Final Answer:

   [1, 2, 3] -> Option B

4. Quick Check:

   Unique elements extracted as [1, 2, 3] [OK]

Hint: Print slice up to i+1 for unique elements [OK]

Common Mistakes:

• Printing full array instead of slice
• Not updating i correctly
• Confusing pointer roles

4. Identify the error in the following code snippet for removing duplicates from a sorted array:

nums = [1, 1, 2, 3, 3]
i = 0
for j in range(1, len(nums)):
    if nums[i] == nums[j]:
        i += 1
        nums[i] = nums[j]
print(nums[:i+1])

medium

A. The condition should check if nums[j] != nums[i]

B. The pointer i should start at 1 instead of 0

C. The loop should start from j = 0

D. The print statement should print the entire array

Solution

1. Step 1: Analyze the if condition

   The code increments i and updates nums[i] only when nums[i] == nums[j], which is incorrect for removing duplicates.

2. Step 2: Correct the condition

   It should increment i and update nums[i] when nums[j] != nums[i] to keep unique elements.

3. Final Answer:

   The condition should check if nums[j] != nums[i] -> Option A

4. Quick Check:

   Use != to detect new unique elements [OK]

Hint: Use != to find unique elements, not == [OK]

Common Mistakes:

• Using == instead of !=
• Starting pointers incorrectly
• Printing wrong slice of array

5. Given a sorted array nums = [0,0,1,1,1,2,2,3,3,4], what will be the array state after applying the two-pointer method to remove duplicates in-place?

hard

A. [0, 1, 2, 3, 4]

B. [0, 0, 1, 2, 3, 4]

C. [0, 1, 1, 2, 3, 4]

D. [0, 1, 2, 2, 3, 4]

Solution

1. Step 1: Apply two-pointer logic on nums

   Start with i=0, j scans from 1 to end. Each time nums[j] != nums[i], increment i and set nums[i] = nums[j].

2. Step 2: Track unique elements placement

   Unique elements placed at start: 0,1,2,3,4. The array's first 5 elements become [0,1,2,3,4].

3. Final Answer:

   [0, 1, 2, 3, 4] -> Option A

4. Quick Check:

   Unique sorted elements remain at front [OK]

Hint: Unique elements stay at front after two-pointer overwrite [OK]

Common Mistakes:

• Assuming duplicates remain in front
• Not updating i correctly
• Confusing array slicing after operation