Remove Duplicates from Sorted Array Two Pointer in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the slow pointer at the start of the array.

DSA C

int slow = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Starting slow pointer at 1 instead of 0.

Using negative or out-of-bound indices.

The slow pointer starts at index 0 to track the position of unique elements.

2fill in blank

medium

Complete the code to move the fast pointer through the array.

DSA C

for (int fast = [1]; fast < n; fast++) {

Drag options to blanks, or click blank then click option'

A-1

B1

Cn

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting fast pointer at 0 causing redundant comparisons.

Starting fast pointer at n causing out-of-bound errors.

The fast pointer starts at 1 to compare with the slow pointer at 0.

3fill in blank

hard

Fix the error in the condition to check if current element is different from last unique element.

DSA C

if (nums[fast] [1] nums[slow]) {

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A>=

B<=

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' causing no duplicates to be removed.

Using '<=' or '>=' causing incorrect comparisons.

We check if the current element is not equal to the last unique element to identify duplicates.

4fill in blank

hard

Fill both blanks to update the slow pointer and assign the new unique element.

DSA C

slow[1];
nums[slow] = nums[2];

Drag options to blanks, or click blank then click option'

A++

B--

C[fast]

D[slow]

Attempts:

3 left

💡 Hint

Common Mistakes

Decrementing slow pointer instead of incrementing.

Assigning nums[slow] to itself instead of nums[fast].

We increment slow by 1 and assign nums[fast] to nums[slow] to keep unique elements.

5fill in blank

hard

Fill in the blanks to return the count of unique elements after processing.

DSA C

return slow [1] [2];

Drag options to blanks, or click blank then click option'

A+

B-

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning slow without adding 1 causing off-by-one error.

Subtracting 1 causing wrong count.

Adding 0 which does not change the value.

We return slow + 1 because slow is index-based, and adding 1 gives the count.

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