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 intact.

🎯 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, 2, 3, 4, 4, 5

Create an integer variable called length and set it to the size of nums

Use two integer pointers i and j to remove duplicates in-place

Print the array after removing duplicates, showing only unique elements separated by spaces

💡 Why This Matters

🌍 Real World

Removing duplicates from sorted data is common in cleaning up lists like contact numbers, product IDs, or timestamps.

💼 Career

Understanding two-pointer techniques and in-place array modification is useful for coding interviews and efficient data processing.

1

Create the sorted array

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

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 the length of the array

Create an integer variable called length and set it to the number of elements in nums using sizeof(nums) / sizeof(nums[0]).

DSA C

int nums[] = {1, 1, 2, 2, 3, 4, 4, 5};
// Create length variable to hold the size of nums
// Your code here

Hint

Use sizeof(nums) / sizeof(nums[0]) to get the number of elements.

3

Remove duplicates using two pointers

Use two integer variables i and j. Initialize i to 0. Use a for loop with j from 1 to length - 1. Inside the loop, if nums[j] is not equal to nums[i], increment i and set nums[i] = nums[j]. After the loop, update length to i + 1.

DSA C

int nums[] = {1, 1, 2, 2, 3, 4, 4, 5};
int length = sizeof(nums) / sizeof(nums[0]);
// Remove duplicates using two pointers i and j
// Your code here

Hint

Use i to track the position of unique elements and j to scan through the array.

4

Print the array after removing duplicates

Use a for loop with variable k from 0 to length - 1. Inside the loop, print nums[k] followed by a space. After the loop, print a newline.

DSA C

int nums[] = {1, 1, 2, 2, 3, 4, 4, 5};
int length = sizeof(nums) / sizeof(nums[0]);
int i = 0;
for (int j = 1; j < length; j++) {
    if (nums[j] != nums[i]) {
        i++;
        nums[i] = nums[j];
    }
}
length = i + 1;
// Print the unique elements in nums
// Your code here

Hint

Use printf("%d ", nums[k]); inside the loop and printf("\n"); after the loop.

Practice

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

easy

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

B. To sort the array before removing duplicates

C. To create a new array with unique elements

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 A

4. Quick Check:

   Two pointers help remove duplicates in-place [OK]

Hint: Two pointers track unique and current elements [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 two pointers i and j for the two-pointer approach in C?

easy

A. int i = 0, j = 1;

B. int i = 1, j = 0;

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

D. int i, j = 0;

Solution

1. Step 1: Understand pointer roles

   Pointer i starts at 0 (first unique), j starts at 1 (scanning next elements).

2. Step 2: Correct initialization syntax

   Both pointers must be initialized properly; int i = 0, j = 1; is correct in C.

3. Final Answer:

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

4. Quick Check:

   Initialize i=0 and j=1 for two-pointer [OK]

Hint: Start i at 0 and j at 1 for scanning [OK]

Common Mistakes:

• Swapping initial values of i and j
• Initializing both pointers to 0
• Incorrect declaration syntax

3. What will be the output length after removing duplicates using the two-pointer method on the sorted array [1, 1, 2, 3, 3, 4]?

medium

A. 6

B. 5

C. 3

D. 4

Solution

1. Step 1: Trace the two-pointer process

   Start with i=0, j=1. Compare elements: duplicates skipped, unique elements moved forward.

2. Step 2: Count unique elements

   Unique elements are 1, 2, 3, 4. So length is 4 after removal.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Unique elements count = 4 [OK]

Hint: Count unique elements after skipping duplicates [OK]

Common Mistakes:

• Counting duplicates as unique
• Assuming length stays same
• Miscounting unique elements

4. Identify the bug in this C code snippet for removing duplicates from a sorted array:

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

medium

A. Does not check for array size zero

B. Pointers i and j are initialized incorrectly

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

D. Uses while loop instead of for loop

Solution

1. Step 1: Analyze return value

   Variable i tracks last unique index, so length is i+1, not i.

2. Step 2: Identify correct return

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

3. Final Answer:

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

4. Quick Check:

   Return length = i+1, not i [OK]

Hint: Return last index + 1 for length [OK]

Common Mistakes:

• Off-by-one error in return value
• Ignoring empty array edge case
• Confusing pointer initialization

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, 0, 1, 1, 2, 2, 3, 3, 4, 4]

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

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

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

Solution

1. Step 1: Apply two-pointer overwrite

   Unique elements 0,1,2,3,4 are moved to front positions in array.

2. Step 2: Final array state

   Positions after unique elements remain unchanged or ignored (represented as underscores).

3. Final Answer:

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

4. Quick Check:

   Array front holds unique elements [OK]

Hint: Unique elements overwrite front, rest ignored [OK]

Common Mistakes:

• Assuming duplicates remain in place
• Misplacing unique elements
• Confusing array length with content