Why Two Pointer Technique Beats Brute Force in DSA Rust - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

let mut left = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Darray.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first element.

Using array length as index causes out-of-bounds error.

The left pointer should start at index 0 to begin scanning from the start of the array.

2fill in blank

medium

Complete the code to move the right pointer to the last element of the array.

DSA Rust

let mut right = [1];

Drag options to blanks, or click blank then click option'

A0

B1

Carray.len() - 1

Darray.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right pointer to array.len() causes index out of range.

Setting right pointer to 0 starts at the beginning, not the end.

The right pointer should start at the last valid index, which is length minus one.

3fill in blank

hard

Fix the error in the loop condition to continue while left is less than right.

DSA Rust

while [1] < right {

Drag options to blanks, or click blank then click option'

Aright

Bright - 1

Cleft + 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right causes the loop to never run.

Using left + 1 < right changes the logic incorrectly.

The loop should run while the left pointer is less than the right pointer to avoid overlap.

4fill in blank

hard

Fill both blanks to correctly move pointers based on sum comparison.

DSA Rust

if sum < target {
    [1] += 1;
} else {
    [2] -= 1;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Csum

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Moving sum or target instead of pointers.

Moving pointers in wrong direction.

If sum is less than target, move left pointer right to increase sum; else move right pointer left to decrease sum.

5fill in blank

hard

Fill all three blanks to create a hashmap of numbers and their indices for the hashmap approach.

DSA Rust

let mut map = std::collections::HashMap::new();
for ([1], &num) in nums.iter().enumerate() {
    if let Some(&idx) = map.get(&([3] - num)) {
        return vec![idx, [1]];
    }
    map.insert(num, [2]);
}

Drag options to blanks, or click blank then click option'

Ai

Ctarget

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names for index or target.

Mixing up num and target in map lookup.

We use 'i' as the index variable, insert num with index i, and check if target - num exists in map.

Practice

1. What is the main advantage of the two pointer technique over brute force in array problems?

easy

A. It always finds the maximum element in the array

B. It reduces time complexity by avoiding nested loops

C. It requires sorting the array first in all cases

D. It uses more memory to store intermediate results

Solution

1. Step 1: Understand brute force approach

   Brute force often uses nested loops, leading to higher time complexity like O(n²).

2. Step 2: Compare with two pointer technique

   Two pointer technique uses two indices moving through the array, reducing complexity to O(n).

3. Final Answer:

   It reduces time complexity by avoiding nested loops -> Option B

4. Quick Check:

   Two pointer = faster than brute force [OK]

Hint: Two pointers move linearly, avoiding double loops [OK]

Common Mistakes:

• Thinking two pointer always needs sorting
• Confusing memory use with speed
• Assuming it finds max element always

2. Which of the following is the correct Rust syntax to initialize two pointers at the start and end of a vector named nums?

easy

A. let (mut left, mut right) = (0, nums.len() - 1);

B. let left = 0; let right = nums.len();

C. let (left, right) = (nums[0], nums[nums.len()]);

D. let mut left = nums.len() - 1; let mut right = 0;

Solution

1. Step 1: Identify pointer initialization

   Two pointers are indices, so left starts at 0 and right at last index (length - 1).

2. Step 2: Check Rust syntax correctness

   let (mut left, mut right) = (0, nums.len() - 1); correctly declares mutable variables with proper tuple assignment and index calculation.

3. Final Answer:

   let (mut left, mut right) = (0, nums.len() - 1); -> Option A

4. Quick Check:

   Correct Rust pointer init = let (mut left, mut right) = (0, nums.len() - 1); [OK]

Hint: Pointers are indices: start=0, end=length-1 [OK]

Common Mistakes:

• Using nums.len() instead of nums.len() - 1 for last index
• Assigning values from vector elements instead of indices
• Not declaring pointers as mutable when needed

3. What will be the output of the following Rust code snippet?

let nums = vec![1, 2, 3, 4, 5];
let mut left = 0;
let mut right = nums.len() - 1;
let mut sum_pairs = Vec::new();
while left < right {
    sum_pairs.push(nums[left] + nums[right]);
    left += 1;
    right -= 1;
}
println!("{:?}", sum_pairs);

medium

A. [5, 6, 7]

B. [6, 7]

C. [5, 7]

D. [6, 6]

Solution

1. Step 1: Trace pointer movements and sums

   Initially left=0, right=4; sums: nums[0]+nums[4]=1+5=6; then left=1, right=3; sums: 2+4=6; loop stops when left=2, right=2.

2. Step 2: Collect sums in vector

   sum_pairs contains [6, 6] after loop ends.

3. Final Answer:

   [6, 6] -> Option D

4. Quick Check:

   Pairs sum = [6, 6] [OK]

Hint: Sum pairs from ends moving inward [OK]

Common Mistakes:

• Including middle element sum when left == right
• Miscalculating indices or sums
• Assuming loop runs when left == right

4. Identify the bug in this Rust code using two pointers to find if a pair sums to 10:

let nums = vec![1, 3, 5, 7, 9];
let mut left = 0;
let mut right = nums.len();
while left < right {
    let sum = nums[left] + nums[right];
    if sum == 10 {
        println!("Found");
        break;
    } else if sum < 10 {
        left += 1;
    } else {
        right -= 1;
    }
}

medium

A. right is initialized to nums.len(), which is out of bounds

B. left pointer should start at nums.len() - 1

C. The loop condition should be left <= right

D. The sum calculation should use nums[left] - nums[right]

Solution

1. Step 1: Check pointer initialization

   right is set to nums.len(), which is one past the last valid index (out of bounds).

2. Step 2: Correct pointer initialization

   right should be nums.len() - 1 to point to last element.

3. Final Answer:

   right is initialized to nums.len(), which is out of bounds -> Option A

4. Quick Check:

   Index out of bounds error = right is initialized to nums.len(), which is out of bounds [OK]

Hint: Last index = length - 1, not length [OK]

Common Mistakes:

• Using length as index causing runtime panic
• Changing loop condition incorrectly
• Misusing sum calculation operator

5. Given a sorted vector nums, how can the two pointer technique be used to find all unique pairs that sum to a target value without duplicates?

hard

A. Sort the vector first, then use a hash map to find complements

B. Use nested loops to check all pairs and store unique pairs in a set

C. Use two pointers from start and end, move inward skipping duplicates after finding a pair

D. Use a single pointer and binary search for complement for each element

Solution

1. Step 1: Initialize two pointers at start and end

   Set left = 0 and right = nums.len() - 1 to scan pairs from both ends.

2. Step 2: Move pointers inward and skip duplicates

   When a pair sums to target, record it, then move pointers skipping over duplicate values to avoid repeated pairs.

3. Final Answer:

   Use two pointers from start and end, move inward skipping duplicates after finding a pair -> Option C

4. Quick Check:

   Two pointer + skip duplicates = unique pairs [OK]

Hint: Skip duplicates after finding pairs to avoid repeats [OK]

Common Mistakes:

• Using nested loops causing slow performance
• Not skipping duplicates leading to repeated pairs
• Using hash map or binary search unnecessarily