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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the second pointer for the two-pointer technique.

DSA Javascript

let i = 0;
let [1] = 1;

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

Aj

Bk

Cm

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same variable name for both pointers.

Choosing a pointer name that is not a single letter.

The second pointer is commonly named 'j' to traverse the array while 'i' tracks the position of unique elements.

2fill in blank

medium

Complete the code to compare elements at the two pointers.

DSA Javascript

if (nums[[1]] !== nums[i]) {

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

Ai

Bk

Cj

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing the element at 'i' with itself.

Using an undefined pointer variable.

We compare the element at the second pointer 'j' with the element at 'i' to find duplicates.

3fill in blank

hard

Fix the error in updating the unique elements count.

DSA Javascript

i[1];

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

A--

B++

C-= 1

D+= 2

Attempts:

3 left

💡 Hint

Common Mistakes

Decrementing the pointer instead of incrementing.

Increasing by 2 which skips positions.

We increment 'i' by 1 to move to the next position for storing a unique element.

4fill in blank

hard

Fill both blanks to assign the new unique element and move the second pointer forward.

DSA Javascript

nums[i] = nums[[1]];
[2]++;

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

Aj

Bi

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing the wrong pointer.

Assigning from the wrong index.

We assign the element at 'j' to position 'i' and then increment 'j' to continue scanning.

5fill in blank

hard

Fill all three blanks to complete the two-pointer loop for removing duplicates.

DSA Javascript

while ([1] < nums.length) {
  if (nums[[2]] !== nums[i]) {
    i++;
    nums[i] = nums[[3]];
  }
  [1]++;
}

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

Aj

Bi

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing 'i' and 'j' in scanning positions.

Incrementing the wrong pointer in the loop.

The pointer 'j' scans through the array, so all blanks related to scanning use 'j'.

Practice

1. What is the main purpose of using the two-pointer technique in removing duplicates from a sorted array?

easy

A. To reverse the array while removing duplicates

B. To sort the array again after duplicates are removed

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

D. To efficiently overwrite duplicates and keep unique elements at the start

Solution

1. Step 1: Understand the two-pointer technique

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

2. Step 2: Purpose of the technique

   It overwrites duplicates by placing unique elements at the front, avoiding extra space.

3. Final Answer:

   To efficiently overwrite duplicates and keep unique elements at the start -> Option D

4. Quick Check:

   Two-pointer = overwrite duplicates [OK]

Hint: Two pointers: one scans, one writes unique values [OK]

Common Mistakes:

• Thinking it sorts the array again
• Confusing counting duplicates with removing them
• Assuming it reverses the array

2. Which of the following is the correct syntax to initialize two pointers for removing duplicates from a sorted array nums in JavaScript?

easy

A. let i = 1, j = 0;

B. let i = 0, j = 1;

C. let i = 0, j = 0;

D. let i = 1, j = 1;

Solution

1. Step 1: Identify pointer roles

   Pointer i tracks last unique element index, j scans from next element.

2. Step 2: Correct initialization

   i starts at 0 (first element), j starts at 1 (second element) to compare.

3. Final Answer:

   let i = 0, j = 1; -> Option B

4. Quick Check:

   i=0, j=1 start positions [OK]

Hint: Start i at 0, j at 1 to compare neighbors [OK]

Common Mistakes:

• Starting both pointers at 0
• Swapping i and j initialization
• Starting both pointers at 1

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

medium

A. [1, 2, 3]

B. [1, 1, 2, 3]

C. [1, 2, 2, 3]

D. [1, 3, 2]

Solution

1. Step 1: Traverse and compare elements

   Start with i=0, j=1. Compare nums[j] with nums[i]. If different, increment i and copy nums[j].

2. Step 2: Update array with unique elements

   After processing, unique elements at start are [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Unique sorted elements = [1, 2, 3] [OK]

Hint: Unique elements stay at front in sorted order [OK]

Common Mistakes:

• Leaving duplicates in the output
• Mixing order of elements
• Not updating the array in place

4. Consider this code snippet for removing duplicates from a sorted array nums. What is the bug?

let i = 0;
for (let j = 1; j < nums.length; j++) {
  if (nums[j] !== nums[i]) {
    i = j;
    nums[i] = nums[j];
  }
}
return i + 1;

medium

A. Pointer i should be incremented, not assigned j

B. The loop should start from j = 0

C. The comparison should be nums[j] === nums[i]

D. Return statement should be i instead of i + 1

Solution

1. Step 1: Analyze pointer update

   i should move forward by incrementing to next unique position, not jump to j.

2. Step 2: Correct pointer increment

   Use i++ to move i forward, then assign nums[i] = nums[j].

3. Final Answer:

   Pointer i should be incremented, not assigned j -> Option A

4. Quick Check:

   i++ to move pointer forward [OK]

Hint: Increment i to place unique elements, don't assign j to i [OK]

Common Mistakes:

• Assigning i = j instead of i++
• Starting loop at wrong index
• Wrong comparison operator

5. Given a sorted array nums = [0,0,1,1,1,2,2,3,3,4], what will be the length of the array after removing duplicates using the two-pointer method, and what will the first part of the array look like?

hard

A. Length: 4, Array start: [0, 1, 2, 3]

B. Length: 6, Array start: [0, 1, 1, 2, 3, 4]

C. Length: 5, Array start: [0, 1, 2, 3, 4]

D. Length: 5, Array start: [0, 0, 1, 2, 3]

Solution

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

   Traverse array, move unique elements forward: 0,1,2,3,4.

2. Step 2: Count unique elements and array start

   There are 5 unique elements, so length is 5, and array start is [0,1,2,3,4].

3. Final Answer:

   Length: 5, Array start: [0, 1, 2, 3, 4] -> Option C

4. Quick Check:

   Unique count = 5, array front = [0,1,2,3,4] [OK]

Hint: Count unique elements, array front holds them [OK]

Common Mistakes:

• Counting duplicates as unique
• Not updating array in place
• Miscounting length after removal