Remove Duplicates from Sorted Array Two Pointer in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the slow pointer at the start of the array.

DSA Rust

let mut slow = [1];

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A0

B1

Carr.len()

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting slow pointer at 1 instead of 0.

Using array length as starting index.

Using negative index which is invalid.

The slow pointer starts at index 0 to track the position of unique elements.

2fill in blank

medium

Complete the code to loop through the array starting from the second element.

DSA Rust

for fast in [1]..arr.len() {

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Aarr.len() - 1

B1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 0 causing redundant comparison.

Starting loop at 2 skipping elements.

Using arr.len() - 1 which misses last element.

The fast pointer starts from index 1 to compare with the slow pointer at index 0.

3fill in blank

hard

Fix the error in the condition to check if current element is different from last unique element.

DSA Rust

if arr[fast] [1] arr[slow] {

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A!=

B==

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using == which checks for equality instead of difference.

Using >= or <= which are not correct for duplicate detection.

We check if the current element is not equal to the last unique element to identify duplicates.

4fill in blank

hard

Fill both blanks to update the slow pointer and assign the new unique element.

DSA Rust

slow = slow [1] 1;
arr[slow] = arr[[2]];

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A+

B-

C*

Dfast

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Assigning arr[slow] to arr[fast] instead of arr[fast] to arr[slow].

We increment slow by 1 and assign the current fast element to arr[slow].

5fill in blank

hard

Fill all three blanks to return the length of unique elements after processing.

DSA Rust

return [1] + [2] - [3];

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Aslow

B1

C0

Dfast

Attempts:

3 left

💡 Hint

Common Mistakes

Returning slow without adding 1.

Using fast instead of slow for length.

Subtracting 1 which is incorrect.

The length of unique elements is slow + 1 because slow is index-based starting at 0.

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