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

The Big Idea

What if you could clean up long sorted lists in just one quick pass without extra space?

The Scenario

Imagine you have a long list of names sorted alphabetically, but some names appear more than once. You want to write down only one copy of each name on a new list.

Doing this by checking every name one by one and comparing it to all others is tiring and slow.

The Problem

Manually checking each name against all others means repeating work many times. It takes a lot of time and you might miss duplicates or make mistakes.

This slow process wastes your energy and can cause errors if the list is very long.

The Solution

The two-pointer method uses two markers moving through the list to quickly find and keep only unique names. It compares neighbors and skips duplicates without extra space.

This way, you get a clean list fast and without confusion.

Before vs After

✗ Before

for (int i = 0; i < arr.length; i++) {
  for (int j = i + 1; j < arr.length; j++) {
    if (arr[i] == arr[j]) {
      // remove duplicate
    }
  }
}

✓ After

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

What It Enables

This method lets you quickly clean sorted lists by removing duplicates in place, saving time and memory.

Real Life Example

When managing a sorted contact list on your phone, this method helps keep only one entry per person, making your list neat and easy to use.

Key Takeaways

Manual duplicate removal is slow and error-prone.

Two-pointer technique efficiently removes duplicates in sorted arrays.

It works in place, saving memory and time.

Practice

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

easy

A. To sort the array before removing duplicates

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

C. To create a new array with unique elements

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

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, keeping unique elements at the start.

3. Final Answer:

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

4. Quick Check:

   Two pointers help remove duplicates without extra space [OK]

Hint: Two pointers move to keep unique elements compact [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 the two pointers i and j for the problem in Java?

easy

A. int i = 0, j = 1;

B. int i = 1, j = 0;

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

D. int i = 1; int j = 1;

Solution

1. Step 1: Identify pointer roles

   Pointer i starts at 0 to track unique elements, j starts at 1 to scan ahead.

2. Step 2: Check initialization correctness

   Initializing i = 0 and j = 1 allows scanning from second element while i marks last unique.

3. Final Answer:

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

4. Quick Check:

   Start i at 0, j at 1 to scan array [OK]

Hint: Start i at 0 and 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 length after removing duplicates using two pointers on the sorted array [1, 1, 2, 2, 3]?

medium

A. 2

B. 3

C. 5

D. 4

Solution

1. Step 1: Trace the two-pointer process

   Start with i=0, j=1; compare elements and overwrite duplicates: unique elements are 1, 2, 3.

2. Step 2: Count unique elements

   After processing, unique elements count is 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Unique elements count = 3 [OK]

Hint: Count unique elements after skipping duplicates [OK]

Common Mistakes:

• Counting duplicates instead of unique
• Assuming length stays same
• Miscounting when duplicates are consecutive

4. Identify the error in the following Java code snippet for removing duplicates from a sorted array:

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

medium

A. Pointers initialized incorrectly

B. Incorrect comparison operator used

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

D. Does not increment j inside the loop

Solution

1. Step 1: Analyze return value

   The variable i tracks last unique index, so length is i + 1.

2. Step 2: Identify mistake in return

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

3. Final Answer:

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

4. Quick Check:

   Length = last index + 1 [OK]

Hint: Return i + 1 for length, not i [OK]

Common Mistakes:

• Returning i instead of i + 1
• Forgetting to increment j
• Using wrong comparison operator

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, 1, 2, 3, 4, _, _, _, _, _]

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

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

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

Solution

1. Step 1: Apply two-pointer overwrite

   Start with i=0, j=1; overwrite duplicates by moving unique elements forward: unique elements are 0,1,2,3,4.

2. Step 2: Final array state

   First five positions hold unique elements [0,1,2,3,4], rest remain unchanged or ignored.

3. Final Answer:

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

4. Quick Check:

   Unique elements compacted at start [OK]

Hint: Unique elements fill front, duplicates ignored after [OK]

Common Mistakes:

• Assuming duplicates remain in place
• Misplacing unique elements
• Counting extra elements beyond unique count