Remove Duplicates from Sorted Array Two Pointer in DSA Rust - 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 at most twice.This is like organizing a row of books where you only want to keep two copies of each title.

🎯 Goal: Build a Rust program that removes extra duplicates from a sorted array, allowing each number to appear at most twice.You will use the two-pointer technique to do this efficiently.

📋 What You'll Learn

Create a vector called nums with the exact values: [1, 1, 1, 2, 2, 3]

Create a variable called max_allowed and set it to 2

Use two pointers named i and j to remove duplicates allowing at most two occurrences

Print the modified vector nums after removing extra duplicates

💡 Why This Matters

🌍 Real World

Removing extra duplicates is useful in data cleaning, like keeping only a few copies of repeated records in a database or list.

💼 Career

This technique is common in software engineering roles that handle data processing, optimization, and memory-efficient algorithms.

1

Create the initial sorted vector

Create a vector called nums with the exact values [1, 1, 1, 2, 2, 3].

DSA Rust

# Create the vector nums with values [1, 1, 1, 2, 2, 3]
# Your code here

Hint

Use let mut nums = vec![1, 1, 1, 2, 2, 3]; to create the vector.

2

Set the maximum allowed duplicates

Create a variable called max_allowed and set it to 2 inside the main function after the vector nums.

DSA Rust

fn main() {
    let mut nums = vec![1, 1, 1, 2, 2, 3];
    // Create max_allowed variable and set it to 2
    // Your code here
}

Hint

Use let max_allowed = 2; to set the limit.

3

Use two pointers to remove extra duplicates

Use two pointers named i and j to iterate over nums and remove duplicates allowing at most max_allowed occurrences. Update nums in place inside the main function.

DSA Rust

    let mut i = 0;
    for j in 0..nums.len() {
        if i < max_allowed || nums[j] != nums[i - max_allowed] {
            nums[i] = nums[j];
            i += 1;
        }
    }
    nums.truncate(i);
    // Print the nums vector

Hint

Use a for loop with j from 0 to nums.len(). Use i to track the position to write. Check if i < max_allowed or current number differs from the number max_allowed positions before.

4

Print the cleaned vector

Print the vector nums after removing extra duplicates using println!("{:?}", nums); inside the main function.

DSA Rust

Hint

Use println!("{:?}", nums); to print the vector.

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