Why Remove Duplicates from Sorted Array Two Pointer in DSA Javascript?

The Big Idea

What if you could clean up long lists in just one quick pass without mistakes?

The Scenario

Imagine you have a long list of names sorted alphabetically, but some names appear more than once. You want to make a clean list with each name only once. Doing this by checking each name one by one and removing duplicates manually is tiring and slow.

The Problem

Manually scanning the list and removing duplicates means you keep shifting names around, which takes a lot of time and can cause mistakes like skipping names or deleting the wrong one. It's easy to get lost and waste effort.

The Solution

The two-pointer method lets you quickly find and keep only unique names by moving through the list just once. One pointer tracks where to place the next unique name, and the other scans through the list. This way, you avoid extra work and errors.

Before vs After

✗ Before

for (let i = 1; i < arr.length; i++) {
  if (arr[i] === arr[i - 1]) {
    arr.splice(i, 1);
    i--;
  }
}

✓ After

let uniqueIndex = 0;
for (let currentIndex = 1; currentIndex < arr.length; currentIndex++) {
  if (arr[currentIndex] !== arr[uniqueIndex]) {
    uniqueIndex++;
    arr[uniqueIndex] = arr[currentIndex];
  }
}
return uniqueIndex + 1;

What It Enables

This method makes it easy to clean sorted lists fast and safely, enabling efficient data processing and saving time.

Real Life Example

When a music app shows your playlist, it removes duplicate songs so you don't see repeats. Using this method, the app quickly cleans the list before showing it to you.

Key Takeaways

Manual removal is slow and error-prone.

Two-pointer method scans once and keeps unique items efficiently.

It helps process sorted data quickly and safely.

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