Remove Duplicates from Sorted Array Two Pointer in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the 'Remove Duplicates from Sorted Array Two Pointer' technique?

To remove duplicate elements from a sorted array in-place, so that each element appears only once, and return the new length of the array.

beginner

Why does the two-pointer approach work efficiently for removing duplicates in a sorted array?

Because the array is sorted, duplicates appear consecutively. One pointer tracks the position to insert unique elements, and the other scans through the array to find new unique elements.

beginner

In the two-pointer method, what do the pointers usually represent?

One pointer (slow) marks the position of the last unique element, and the other pointer (fast) scans ahead to find the next unique element.

beginner

What is the time complexity of removing duplicates from a sorted array using the two-pointer technique?

O(n), where n is the number of elements in the array, because each element is visited at most once.

beginner

What does 'in-place' mean in the context of removing duplicates from an array?

It means modifying the original array without using extra space for another array, except for a few variables.

What is the initial position of the slow pointer in the two-pointer approach for removing duplicates?

A 0

B 1

C array length - 1

D array length

The slow pointer starts at index 0, pointing to the first unique element.

When do we move the slow pointer forward in the two-pointer method?

A At every iteration regardless

B When the fast pointer finds a duplicate

C Only at the end

D When the fast pointer finds a new unique element

The slow pointer moves forward only when the fast pointer finds a new unique element to place.

What is the space complexity of the two-pointer approach for removing duplicates from a sorted array?

A O(1)

B O(n)

C O(log n)

D O(n^2)

The approach uses constant extra space, modifying the array in-place.

What happens if the input array is empty?

A Return -1 as error

B Return 1 as the length

C Return 0 as the length

D Crash

If the array is empty, there are no elements, so the length after removing duplicates is 0.

Which of these arrays after removing duplicates using two pointers will have length 3?

A [1, 2, 3, 4]

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

C [1, 1, 1, 1]

D []

After removing duplicates, [1, 1, 2, 3, 3] becomes [1, 2, 3], length 3.

Explain how the two-pointer technique removes duplicates from a sorted array in-place.

Think about how two pointers move and interact in the array.

Describe the advantages of using the two-pointer approach for removing duplicates in a sorted array.

Focus on time, space, and why sorting helps.

Practice

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

easy

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

B. To sort the array before removing duplicates

C. To create a new array with unique elements

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

Solution

1. Step 1: Understand the two-pointer approach

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

2. Step 2: Purpose of pointers in-place

   This method avoids extra space by overwriting duplicates directly in the original array.

3. Final Answer:

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

4. Quick Check:

   Two pointers help remove duplicates in-place [OK]

Hint: Two pointers track unique and current elements [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 C?

easy

A. int i = 0, j = 1;

B. int i = 1, j = 0;

C. int i = 0; int j = 0;

D. int i, j = 0;

Solution

1. Step 1: Understand pointer roles

   Pointer i starts at 0 (first unique), j starts at 1 (scanning next elements).

2. Step 2: Correct initialization syntax

   Both pointers must be initialized properly; int i = 0, j = 1; is correct in C.

3. Final Answer:

   int i = 0, j = 1; -> Option A

4. Quick Check:

   Initialize i=0 and j=1 for two-pointer [OK]

Hint: Start i at 0 and j at 1 for scanning [OK]

Common Mistakes:

• Swapping initial values of i and j
• Initializing both pointers to 0
• Incorrect declaration syntax

3. What will be the output length after removing duplicates using the two-pointer method on the sorted array [1, 1, 2, 3, 3, 4]?

medium

A. 6

B. 5

C. 3

D. 4

Solution

1. Step 1: Trace the two-pointer process

   Start with i=0, j=1. Compare elements: duplicates skipped, unique elements moved forward.

2. Step 2: Count unique elements

   Unique elements are 1, 2, 3, 4. So length is 4 after removal.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Unique elements count = 4 [OK]

Hint: Count unique elements after skipping duplicates [OK]

Common Mistakes:

• Counting duplicates as unique
• Assuming length stays same
• Miscounting unique elements

4. Identify the bug in this C code snippet for removing duplicates from a sorted array:

int removeDuplicates(int* nums, int numsSize) {
    int i = 0, j = 1;
    while (j < numsSize) {
        if (nums[i] != nums[j]) {
            i++;
            nums[i] = nums[j];
        }
        j++;
    }
    return i;
}

medium

A. Does not check for array size zero

B. Pointers i and j are initialized incorrectly

C. Returns i instead of i+1 as new length

D. Uses while loop instead of for loop

Solution

1. Step 1: Analyze return value

   Variable i tracks last unique index, so length is i+1, not i.

2. Step 2: Identify correct return

   Returning i misses counting the first element, causing off-by-one error.

3. Final Answer:

   Returns i instead of i+1 as new length -> Option C

4. Quick Check:

   Return length = i+1, not i [OK]

Hint: Return last index + 1 for length [OK]

Common Mistakes:

• Off-by-one error in return value
• Ignoring empty array edge case
• Confusing pointer initialization

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?

hard

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

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

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

D. [0, 1, 2, 3, 4, 5, _, _, _, _]

Solution

1. Step 1: Apply two-pointer overwrite

   Unique elements 0,1,2,3,4 are moved to front positions in array.

2. Step 2: Final array state

   Positions after unique elements remain unchanged or ignored (represented as underscores).

3. Final Answer:

   [0, 1, 2, 3, 4, _, _, _, _, _] -> Option B

4. Quick Check:

   Array front holds unique elements [OK]

Hint: Unique elements overwrite front, rest ignored [OK]

Common Mistakes:

• Assuming duplicates remain in place
• Misplacing unique elements
• Confusing array length with content