Remove Duplicates from Sorted Array Two Pointer in DSA Python - Time & Space Complexity

Time Complexity: Remove Duplicates from Sorted Array Two Pointer

O(n)

Understanding Time Complexity

We want to know how fast the two-pointer method removes duplicates from a sorted array.

How does the time needed grow as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

    def remove_duplicates(nums):
        if not nums:
            return 0
        i = 0
        for j in range(1, len(nums)):
            if nums[j] != nums[i]:
                i += 1
                nums[i] = nums[j]
        return i + 1
    

This code removes duplicates in-place from a sorted array using two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop that goes through the array once.
• How many times: The loop runs once for each element after the first.

How Execution Grows With Input

As the array size grows, the loop runs more times, directly proportional to the size.

Input Size (n)	Approx. Operations
10	About 9 checks and possible assignments
100	About 99 checks and possible assignments
1000	About 999 checks and possible assignments

Pattern observation: The number of operations grows linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows in a straight line as the array gets bigger.

Common Mistake

[X] Wrong: "Because there are two pointers, the time is doubled or squared."

[OK] Correct: Both pointers move forward through the array only once, so the total work is still proportional to the array size, not squared.

Interview Connect

Understanding this linear time approach shows you can efficiently handle arrays without extra space, a common skill interviewers look for.

Self-Check

"What if the array was not sorted? How would the time complexity change if we still tried to remove duplicates in-place?"

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