Sliding Window Maximum Using Deque in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new empty deque.

DSA Rust

let mut deq: std::collections::VecDeque<usize> = std::collections::VecDeque::[1]();

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

Adefault

Bwith_capacity

Cnew

Dfrom

Attempts:

3 left

💡 Hint

Common Mistakes

Using `with_capacity` requires a size argument.

Using `from` expects an existing collection.

Using `default` is valid but less explicit here.

The method `new()` creates an empty VecDeque.

2fill in blank

medium

Complete the code to remove indices from the back of the deque while the current number is greater.

DSA Rust

while let Some(&back) = deq.back() {
    if nums[back] < nums[i] {
        deq.[1]();
    } else {
        break;
    }
}

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

Apop_front

Bpush_front

Cpush_back

Dpop_back

Attempts:

3 left

💡 Hint

Common Mistakes

Using `pop_front()` removes from the front, which is incorrect here.

Using `push_back()` or `push_front()` adds elements instead of removing.

We remove from the back of the deque using `pop_back()` to maintain decreasing order.

3fill in blank

hard

Fix the error in removing indices from the front of the deque when they are out of the current window.

DSA Rust

if let Some(&front) = deq.front() {
    if i [1] front + k {
        deq.pop_front();
    }
}

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A>

B>=

C<

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using `<` or `>` instead of `>=` causes incorrect removals.

Not comparing with `i - k` but with `i + k`.

We remove from front if the index is out of the window, i.e., if front <= i - k, so the operator is `>=` when rearranged.

4fill in blank

hard

Fill both blanks to push the current index and add the maximum to the result vector.

DSA Rust

deq.[1](i);
if i + 1 >= k {
    result.push(nums[*deq.[2]().unwrap()].clone());
}

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Apush_back

Bpop_front

Cfront

Dback

Attempts:

3 left

💡 Hint

Common Mistakes

Using `pop_front` instead of `front` to access the max index.

Using `back` to get max index, which is incorrect.

We add the current index at the back with `push_back`, and the max is at the front of the deque accessed by `front()`.

5fill in blank

hard

Fill all three blanks to complete the sliding window maximum function using a deque.

DSA Rust

pub fn max_sliding_window(nums: Vec<i32>, k: usize) -> Vec<i32> {
    use std::collections::VecDeque;
    let mut deq: VecDeque<usize> = VecDeque::[1]();
    let mut result = Vec::new();
    for i in 0..nums.len() {
        while let Some(&back) = deq.back() {
            if nums[back] < nums[i] {
                deq.[2]();
            } else {
                break;
            }
        }
        deq.push_back(i);
        if let Some(&front) = deq.front() {
            if front + k <= i {
                deq.[3]();
            }
        }
        if i + 1 >= k {
            result.push(nums[*deq.front().unwrap()]);
        }
    }
    result
}

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

Anew

Bpop_back

Cpop_front

Dpush_back

Attempts:

3 left

💡 Hint

Common Mistakes

Using `push_back` instead of `pop_back` to remove elements.

Using `pop_front` incorrectly on the back side.

Not initializing the deque properly.

We create a new deque with `new()`, remove smaller elements from back with `pop_back()`, and remove out-of-window indices from front with `pop_front()`.

Practice

1. For nums = [1, 3, -1, -3, 5, 3, 6, 7] and k=3, what is the maximum of the first window [1,3,-1]?

easy

A. 1

B. -1

C. 3

D. 5

Solution

1. Step 1: First window

   Window [1, 3, -1] contains elements at indices 0, 1, 2.

2. Step 2: Maximum

   max(1, 3, -1) = 3.

3. Step 3: Deque state

   After processing first window: deque = [1] (index of 3, since 3 is largest and -1 is removed because 3 > -1 and 3 came before -1).

4. Final Answer:

   3 -> Option C

5. Quick Check:

   max(1,3,-1)=3. [OK]

Hint: max(1,3,-1)=3. Deque front holds index 1 (value 3) after processing window. [OK]

Common Mistakes:

• Returning the last element -1
• Returning the first element 1
• Not recognizing 3 as the maximum

2. In the sliding window maximum problem, what does the deque store?

medium

A. The maximum values seen so far

B. Indices of elements in decreasing order of their values

C. All elements in the current window

D. The sum of the current window

Solution

1. Step 1: Deque content

   The deque stores INDICES (not values), arranged so that nums[deque.front()] is always the maximum of the current window.

2. Step 2: Decreasing order

   Elements are kept in decreasing order of their values - smaller elements behind a larger one are useless (they can never be the maximum while the larger element is in the window).

3. Final Answer:

   Indices of elements in decreasing value order. -> Option B

4. Quick Check:

   Deque = indices in decreasing value order. Front = current window max. [OK]

Hint: Deque stores indices, decreasing by value. Front index = current max. [OK]

Common Mistakes:

• Thinking deque stores values not indices
• Thinking all window elements are stored
• Confusing with a max-heap approach

3. What is the time complexity of sliding window maximum using a monotonic deque?

medium

A. O(n²)

B. O(n log n)

C. O(n)

D. O(n × k)

Solution

1. Step 1: Each element pushed once

   Each element is pushed to the deque exactly once.

2. Step 2: Each element popped at most once

   Each element is popped from either front or back at most once. Total push+pop operations: 2n = O(n).

3. Step 3: Overall

   n iterations × O(1) amortized per iteration = O(n).

4. Final Answer:

   O(n) - each element pushed and popped at most once. -> Option C

5. Quick Check:

   Each element: 1 push + at most 1 pop = O(n) total. [OK]

Hint: Monotonic deque: each element pushed once, popped once = O(n). Better than O(n×k) naive. [OK]

Common Mistakes:

• Saying O(n×k) by not accounting for amortized analysis
• Saying O(n log n) confusing with heap-based approach
• Not knowing each element is pushed/popped at most once total

4. When processing a new element i in sliding window maximum, when do we pop from the BACK of the deque?

hard

A. When deque.back() index is outside the window

B. When nums[deque.back()] <= nums[i] - smaller elements are useless

C. When the deque is full (size == k)

D. When nums[i] is negative

Solution

1. Step 1: Why pop from back

   If nums[deque.back()] <= nums[i], the back element can never be the window maximum while nums[i] is in the window. nums[i] is larger AND will leave the window later.

2. Step 2: Keep decreasing order

   Pop all elements from back where nums[back] <= nums[i], then push i. This maintains the monotonic decreasing invariant.

3. Final Answer:

   Pop back when nums[deque.back()] <= nums[i]. -> Option B

4. Quick Check:

   Back pop: current element >= back element → back is useless, remove it. [OK]

Hint: Pop back when nums[back] <= nums[current]. Larger current makes smaller back useless forever. [OK]

Common Mistakes:

• Confusing front pop (out-of-window check) with back pop (useless element removal)
• Popping only when strictly less than, missing equals case
• Thinking deque size triggers back pop

5. When processing a new element i in sliding window maximum, when do we pop from the FRONT of the deque?

hard

A. When nums[deque.front()] < nums[i]

B. When deque.front() <= i - k (front index is outside the current window)

C. When the deque has more than k elements

D. When nums[i] is the new maximum

Solution

1. Step 1: Window boundary

   The current window is [i-k+1, i]. The front of deque holds the maximum's index. If that index <= i-k, it's outside the window.

2. Step 2: Pop expired maximum

   Pop front when deque.front() <= i - k. The front is no longer in the current window - the next front becomes the new maximum.

3. Final Answer:

   Pop front when front index <= i - k (out of window). -> Option B

4. Quick Check:

   Front pop: index out of window (front <= i-k). Back pop: value useless (back <= current). [OK]

Hint: Front pop = expired (index out of window). Back pop = useless (value smaller than current). [OK]

Common Mistakes:

• Confusing front pop condition with back pop condition
• Using < i-k instead of <= i-k (off by one)
• Thinking front pop is based on value comparison