Remove Duplicates from Sorted Array Two Pointer in DSA C# - Build from Scratch

Remove Duplicates from Sorted Array Two Pointer

📖 Scenario: You have a sorted list of numbers that may contain duplicates. You want to clean it up so that each number appears only once, keeping the order.

🎯 Goal: Build a program that removes duplicates from a sorted array using the two-pointer technique and prints the cleaned array.

📋 What You'll Learn

Create an integer array called nums with the exact values: 1, 1, 2, 3, 3, 4, 5, 5, 6

Create an integer variable called writeIndex and set it to 1

Use a for loop with variable readIndex starting from 1 to the end of nums

Inside the loop, compare nums[readIndex] with nums[writeIndex - 1]

If they are different, assign nums[writeIndex] = nums[readIndex] and increment writeIndex

After the loop, print the elements of nums from index 0 to writeIndex - 1 separated by spaces

💡 Why This Matters

🌍 Real World

Removing duplicates from sorted data is common in data cleaning, preparing lists for reports, or optimizing storage.

💼 Career

Understanding two-pointer techniques helps in coding interviews and writing efficient algorithms for array manipulation.

1

Create the sorted array with duplicates

Create an integer array called nums with these exact values: 1, 1, 2, 3, 3, 4, 5, 5, 6

DSA C#

// Create the array nums with the given values
// Your code here

Hint

Use int[] nums = { ... }; to create the array with the exact numbers.

2

Set up the write index

Create an integer variable called writeIndex and set it to 1

DSA C#

int[] nums = {1, 1, 2, 3, 3, 4, 5, 5, 6};
// Create writeIndex and set it to 1
// Your code here

Hint

Use int writeIndex = 1; to start the write pointer at position 1.

3

Use a for loop to remove duplicates

Use a for loop with variable readIndex starting from 1 to the end of nums. Inside the loop, compare nums[readIndex] with nums[writeIndex - 1]. If they are different, assign nums[writeIndex] = nums[readIndex] and increment writeIndex

DSA C#

int[] nums = {1, 1, 2, 3, 3, 4, 5, 5, 6};
int writeIndex = 1;
// Add the for loop and logic to remove duplicates
// Your code here

Hint

Use a for loop from 1 to nums.Length. Compare current and previous unique values. Update and move writeIndex when a new unique number is found.

4

Print the cleaned array

Print the elements of nums from index 0 to writeIndex - 1 separated by spaces

DSA C#

int[] nums = {1, 1, 2, 3, 3, 4, 5, 5, 6};
int writeIndex = 1;
for (int readIndex = 1; readIndex < nums.Length; readIndex++)
{
    if (nums[readIndex] != nums[writeIndex - 1])
    {
        nums[writeIndex] = nums[readIndex];
        writeIndex++;
    }
}
// Print the unique elements separated by spaces
// Your code here

Hint

Use a for loop from 0 to writeIndex - 1 and print each element followed by a space.

Practice

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

easy

A. To create a new array with unique elements

B. To sort the array before removing duplicates

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

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 C

4. Quick Check:

   Two pointers help remove duplicates in-place [OK]

Hint: Two pointers track and overwrite duplicates without extra space [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 C# syntax to initialize the two pointers i and j for the problem?

easy

A. int i = 1; int j = 1;

B. int i = 1, j = 0;

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

D. int i = 0, j = 1;

Solution

1. Step 1: Identify pointer roles

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

2. Step 2: Check initialization correctness

   Initializing i = 0 and j = 1 matches the logic of comparing current and next elements.

3. Final Answer:

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

4. Quick Check:

   Start i at 0 and j at 1 [OK]

Hint: Start i at 0 and j at 1 to compare adjacent elements [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 and new length after running the two-pointer method on the sorted array [1, 1, 2, 3, 3]?

medium

A. [1, 2, 3, 3], length = 4

B. [1, 2, 3], length = 3

C. [1, 1, 2, 3, 3], length = 5

D. [1, 2, 2, 3], length = 4

Solution

1. Step 1: Trace the two-pointer process

   Start with i=0, j=1. Compare elements: skip duplicates, overwrite unique elements at i+1.

2. Step 2: Final array and length

   Unique elements are 1, 2, 3 placed at start. Length is 3.

3. Final Answer:

   [1, 2, 3], length = 3 -> Option B

4. Quick Check:

   Unique elements count = 3 [OK]

Hint: Overwrite duplicates, count unique elements for length [OK]

Common Mistakes:

• Not updating length after removal
• Leaving duplicates in output
• Misplacing unique elements

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

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

medium

A. The function should return i + 1 instead of i

B. Pointers i and j should both start at 1

C. The while loop condition should be j <= nums.Length

D. The array should be sorted inside the function

Solution

1. Step 1: Understand pointer meaning

   Pointer i tracks last unique element index, so length is i + 1.

2. Step 2: Check return value

   Returning i alone misses counting the first element, so return i + 1.

3. Final Answer:

   The function should return i + 1 instead of i -> Option A

4. Quick Check:

   Return length as i + 1 [OK]

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

Common Mistakes:

• Returning i instead of i + 1
• Incorrect loop boundary causing out-of-range
• Sorting array inside function unnecessarily

5. Given a sorted array nums = [0,0,1,1,1,2,2,3,3,4], what will be the array state and returned length after applying the two-pointer duplicate removal method?

hard

A. [0, 1, 2, 3, 4], length = 5

B. [0, 0, 1, 2, 3, 4], length = 6

C. [0, 1, 1, 2, 3, 4], length = 6

D. [0, 1, 2, 3, 3, 4], length = 6

Solution

1. Step 1: Apply two-pointer method step-by-step

   Start with i=0, j=1. Move j forward, overwrite duplicates by placing unique elements at i+1.

2. Step 2: Final array and length

   Unique elements are 0,1,2,3,4 placed at start. Length is 5.

3. Final Answer:

   [0, 1, 2, 3, 4], length = 5 -> Option A

4. Quick Check:

   Unique count = 5 [OK]

Hint: Overwrite duplicates, count unique elements for final length [OK]

Common Mistakes:

• Leaving duplicates in the array
• Returning wrong length
• Misplacing unique elements during overwrite