Remove Duplicates from Sorted Array Two Pointer in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Duplicate Removal

What is the output of the following Rust code that removes duplicates from a sorted array using two pointers?

DSA Rust

fn remove_duplicates(nums: &mut Vec<i32>) -> usize {
    if nums.is_empty() {
        return 0;
    }
    let mut i = 0;
    for j in 1..nums.len() {
        if nums[j] != nums[i] {
            i += 1;
            nums[i] = nums[j];
        }
    }
    i + 1
}

fn main() {
    let mut nums = vec![1,1,2,2,3,4,4,5];
    let len = remove_duplicates(&mut nums);
    nums.truncate(len);
    println!("{:?}", nums);
}

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 how duplicates are overwritten.

The code uses two pointers i and j. i tracks the position of the last unique element. j scans through the array. When nums[j] differs from nums[i], i increments and nums[i] is updated. Finally, the array is truncated to length i+1, removing duplicates.

🧠 Conceptual

intermediate

Understanding Two Pointer Technique

Why does the two pointer technique work efficiently for removing duplicates from a sorted array?

A Because it sorts the array first to group duplicates together.

B Because it uses extra memory to store unique elements separately.

C Because the array is sorted, duplicates are adjacent, so one pointer can track unique elements while the other scans ahead.

D Because it removes duplicates by swapping elements randomly.

💡 Hint

Think about how sorting affects duplicates placement.

Since the array is sorted, duplicates appear next to each other. One pointer keeps track of the last unique element, and the other scans forward to find new unique elements, making the process efficient and in-place.

❓ Predict Output

advanced

Output with All Elements Identical

What will be the output of this Rust code when all elements in the array are identical?

DSA Rust

fn remove_duplicates(nums: &mut Vec<i32>) -> usize {
    if nums.is_empty() {
        return 0;
    }
    let mut i = 0;
    for j in 1..nums.len() {
        if nums[j] != nums[i] {
            i += 1;
            nums[i] = nums[j];
        }
    }
    i + 1
}

fn main() {
    let mut nums = vec![7,7,7,7,7];
    let len = remove_duplicates(&mut nums);
    nums.truncate(len);
    println!("{:?}", nums);
}

A [7]

B [7, 7, 7, 7, 7]

C []

D [7, 7]

💡 Hint

Consider how many unique elements exist.

Since all elements are the same, only one unique element remains after removal. The vector is truncated to length 1, so output is [7].

🔧 Debug

advanced

Identify the Bug in Duplicate Removal Code

What error will this Rust code produce when trying to remove duplicates from a sorted array?

DSA Rust

fn remove_duplicates(nums: &mut Vec<i32>) -> usize {
    let mut i = 0;
    for j in 1..nums.len() {
        if nums[j] != nums[i] {
            i += 1;
            nums[i] = nums[j];
        }
    }
    i + 1
}

fn main() {
    let mut nums = vec![];
    let len = remove_duplicates(&mut nums);
    nums.truncate(len);
    println!("{:?}", nums);
}

A infinite loop

B prints [] with no error

C compilation error due to missing return

D panic at index 0 out of bounds

💡 Hint

Check what happens when the input vector is empty.

The code lacks an empty vector check. For empty input, the loop (1..0) does not run, but nums[0] is accessed when i=0 and returned as i+1, which is 1. However, since nums is empty, accessing nums[0] in the loop condition or elsewhere causes a panic at index 0 out of bounds.

🚀 Application

expert

Length of Array After Removing Duplicates

Given the Rust function below, what is the length of the array after removing duplicates from this input: [0,0,1,1,1,2,2,3,3,4]?

DSA Rust

fn remove_duplicates(nums: &mut Vec<i32>) -> usize {
    if nums.is_empty() {
        return 0;
    }
    let mut i = 0;
    for j in 1..nums.len() {
        if nums[j] != nums[i] {
            i += 1;
            nums[i] = nums[j];
        }
    }
    i + 1
}

A 4

B 5

C 10

D 6

💡 Hint

Count unique elements in the input array.

The unique elements are 0,1,2,3,4 which count to 5. The function returns the count of unique elements after removing duplicates.

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 efficiently track unique elements and overwrite duplicates in-place

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 B

4. Quick Check:

   Two pointers help remove duplicates in-place [OK]

Hint: Two pointers move to overwrite duplicates without extra space [OK]

Common Mistakes:

• Thinking two pointers create a new array
• Assuming sorting is needed again
• Counting duplicates instead of removing

2. Which of the following Rust code snippets correctly initializes two pointers for the Remove Duplicates problem?

easy

A. let mut i = 1; let mut j = 0;

B. let mut i = 0; let mut j = 1;

C. let i = 1; let j = 0;

D. let i = 0; let j = 0;

Solution

1. Step 1: Identify pointer roles

   Pointer i tracks last unique index, starting at 0; j scans from 1.

2. Step 2: Match initialization

   Correct initialization is let mut i = 1; let mut j = 0; to start comparing from second element.

3. Final Answer:

   let mut i = 1; let mut j = 0; -> Option A

4. Quick Check:

   i=1, j=0 starts scanning correctly [OK]

Hint: Start i at 1 and j at 0 to compare adjacent elements [OK]

Common Mistakes:

• Swapping i and j initialization
• Not using mutable variables
• Starting both pointers at 0

3. What will be the content of the array after running the two-pointer duplicate removal on let mut nums = vec![1, 1, 2, 3, 3];?

medium

A. [1, 2, 3, 3, 3]

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

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

D. [1, 2, 3, 0, 0]

Solution

1. Step 1: Apply two-pointer logic

   Start with i=0, j=1; compare nums[j] with nums[i]. When different, increment i and copy nums[j].

2. Step 2: Trace array changes

   nums changes to [1, 2, 3, 3, 3], but only first 3 elements are unique; rest are leftover values.

3. Final Answer:

   [1, 2, 3, _, _] -> Option C

4. Quick Check:

   Unique elements at start, duplicates ignored [OK]

Hint: Unique elements fill front; leftover values remain unchanged [OK]

Common Mistakes:

• Assuming duplicates are removed from array length
• Expecting zeros instead of leftover values
• Not understanding in-place overwrite

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

let mut i = 0;
for j in 1..nums.len() {
    if nums[i] == nums[j] {
        i += 1;
        nums[i] = nums[j];
    }
}

medium

A. Pointer i should not be incremented inside the if block

B. The loop range should start from 0, not 1

C. nums[i] should be compared with nums[i-1], not nums[j]

D. The condition should check for inequality, not equality

Solution

1. Step 1: Analyze the if condition

   The code increments i and copies nums[j] only when nums[i] == nums[j], which is wrong.

2. Step 2: Correct condition logic

   We want to copy nums[j] only when nums[j] != nums[i], meaning a new unique element found.

3. Final Answer:

   The condition should check for inequality, not equality -> Option D

4. Quick Check:

   Copy unique elements when different [OK]

Hint: Copy when elements differ, not when equal [OK]

Common Mistakes:

• Incrementing i on duplicates
• Wrong comparison operator
• Loop starting index confusion

5. Given a sorted vector nums with possible duplicates, how would you modify the two-pointer approach to return the count of unique elements and also ensure the first part of the vector contains only these unique elements in Rust?

hard

A. Use two pointers i=0, j=1; increment i and copy nums[j] when nums[j] != nums[i]; return i+1 as count

B. Use a HashSet to collect unique elements, then overwrite nums with the set; return set size

C. Sort the vector again after removing duplicates; return length of sorted vector

D. Iterate with one pointer, remove duplicates by shifting elements left; return original length minus duplicates

Solution

1. Step 1: Initialize pointers and iterate

   Set i=0 for last unique index, j=1 to scan; compare nums[j] with nums[i].

2. Step 2: Copy unique elements and count

   If nums[j] != nums[i], increment i and copy nums[j] to nums[i]. After loop, unique count is i+1.

3. Final Answer:

   Use two pointers i=0, j=1; increment i and copy nums[j] when nums[j] != nums[i]; return i+1 as count -> Option A

4. Quick Check:

   Two-pointer overwrite returns unique count [OK]

Hint: Increment i only on new unique element, return i+1 [OK]

Common Mistakes:

• Using extra space like HashSet unnecessarily
• Re-sorting after removal
• Incorrect count calculation