Remove Duplicates from Sorted Array Two Pointer in DSA C# - 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 this sorted array?

DSA C#

int[] nums = {1, 1, 2, 2, 3, 4, 4, 5};
int k = RemoveDuplicates(nums);
Console.WriteLine(string.Join(", ", nums[..k]));

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

A 1, 2, 3, 4, 5

B 1, 1, 2, 2, 3, 4, 4, 5

C 1, 2, 2, 3, 4, 5

D 1, 2, 3, 4

💡 Hint

Focus on how the two pointers move and overwrite duplicates.

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

🧠 Conceptual

intermediate

Understanding Two Pointer Roles

In the two-pointer approach to remove duplicates from a sorted array, what is the main role of the pointer 'i'?

A Scans through the array to find duplicates

B Tracks the position to place the next unique element

C Counts the total number of duplicates found

D Points to the last duplicate element

💡 Hint

Think about which pointer moves only when a new unique element is found.

Pointer 'i' marks the index where the next unique element should be placed, effectively building the array of unique elements at the start.

🔧 Debug

advanced

Identify the Bug in Duplicate Removal Code

What error will this code produce when removing duplicates from a sorted array?

DSA C#

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

A NullReferenceException

B Infinite loop

C No error, returns correct length

D IndexOutOfRangeException

💡 Hint

Check the loop boundary conditions carefully.

The loop uses 'j <= nums.Length' which tries to access an index equal to the array length, causing an IndexOutOfRangeException.

❓ Predict Output

advanced

Result Length After Removing Duplicates

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

DSA C#

int[] nums = {0,0,1,1,1,2,2,3,3,4};
int length = RemoveDuplicates(nums);
Console.WriteLine(length);

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

A 5

B 6

C 10

D 4

💡 Hint

Count unique numbers in the array.

The unique numbers are 0, 1, 2, 3, 4, so the length returned is 5.

🚀 Application

expert

Modify Two Pointer to Allow At Most Two Duplicates

How would you modify the two-pointer approach to allow each element to appear at most twice in the sorted array? Which code snippet correctly implements this?

A

int RemoveDuplicates(int[] nums) {
    int i = 1;
    for (int j = 2; j < nums.Length; j++) {
        if (nums[j] != nums[i]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

B

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

C

int RemoveDuplicates(int[] nums) {
    int i = 2;
    for (int j = 2; j < nums.Length; j++) {
        if (nums[j] != nums[i - 2]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

D

int RemoveDuplicates(int[] nums) {
    int i = 1;
    for (int j = 2; j < nums.Length; j++) {
        if (nums[j] != nums[i - 1]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

💡 Hint

Allow duplicates only if the current element is different from the element two positions before.

To allow at most two duplicates, compare current element with the element two positions before. Increment 'i' only if different.

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