Remove Duplicates from Sorted Array Two Pointer in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Duplicate Removal

What is the output array after removing duplicates using the two-pointer method on the sorted array?

DSA Javascript

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

const input = [1,1,2,2,3,4,4,5];
console.log(removeDuplicates(input));

A [1, 2, 3, 4]

B [1, 1, 2, 3, 4, 5]

C [1, 2, 2, 3, 4, 5]

D [1, 2, 3, 4, 5]

💡 Hint

Focus on how the two pointers move and how duplicates are skipped.

The two-pointer method moves through the array, copying unique elements forward. The final array slice contains only unique values.

❓ Predict Output

intermediate

Length of Array After Duplicate Removal

What is the length of the array after removing duplicates using the two-pointer method on this sorted array?

DSA Javascript

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

const input = [0,0,1,1,1,2,2,3,3,4];
console.log(removeDuplicates(input));

A 4

B 6

C 5

D 7

💡 Hint

Count how many unique numbers are in the array.

The unique numbers are 0,1,2,3,4 so length is 5. The slow pointer i advances to 4 after placing the 5th unique element (0 at 0,1 at1,2 at2,3 at3,4 at4), returning i+1=5.

🔧 Debug

advanced

Identify the Error in Two Pointer Duplicate Removal

What error does this code produce when run?

DSA Javascript

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

const input = [1,1,2];
console.log(removeDuplicates(input));

A TypeError: Cannot read property 'undefined' of undefined

B [1, 2]

C [1, 1, 2]

D SyntaxError

💡 Hint

Check the loop boundary condition carefully.

The loop runs with j <= nums.length, so nums[j] is undefined on last iteration causing TypeError.

❓ Predict Output

advanced

Final Array State After Duplicate Removal

What is the final state of the array after running this two-pointer duplicate removal code?

DSA Javascript

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

const input = [2,2,2,3,3,4,5,5];
console.log(removeDuplicates(input));

A [2, 3, 4, 5, 3, 4, 5, 5]

B [2, 3, 4, 5, 5, 5, 5, 5]

C [2, 2, 2, 3, 3, 4, 5, 5]

D [2, 3, 3, 4, 5, 5, 5, 5]

💡 Hint

Only the first part of the array is overwritten with unique values.

The unique elements are moved to the front: 2,3,4,5. The rest of the array remains unchanged.

🧠 Conceptual

expert

Why Two Pointer Method Works for Sorted Arrays

Why does the two-pointer method effectively remove duplicates only when the array is sorted?

A Because the method sorts the array internally before removing duplicates.

B Because duplicates are adjacent, so comparing current and previous elements detects duplicates easily.

C Because it uses a hash map to track duplicates which only works on sorted arrays.

D Because it removes duplicates by swapping elements randomly.

💡 Hint

Think about how duplicates appear in sorted arrays.

In sorted arrays, duplicates are next to each other, so comparing neighbors detects duplicates efficiently.

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