Sliding Window Maximum Using Deque in DSA Rust - Time & Space Complexity

Time Complexity: Sliding Window Maximum Using Deque

O(n)

Understanding Time Complexity

We want to understand how fast the sliding window maximum algorithm runs when using a deque.

Specifically, how the number of steps grows as the input array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the maximum value in every window of size k as it slides over the array nums.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over all elements of nums.
• How many times: Exactly once for each element (n times).
• Inside the loop, elements are added and removed from the deque at most once each.

How Execution Grows With Input

As the input size n grows, each element is processed a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 operations (each element pushed and popped once)
100	About 200 operations
1000	About 2000 operations

Pattern observation: The operations grow roughly linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the algorithm runs in time proportional to the number of elements, processing each element a constant number of times.

Common Mistake

[X] Wrong: "Because there is a nested while loop inside the for loop, the time complexity must be O(n²)."

[OK] Correct: Each element is added and removed from the deque only once, so the inner loop does not run n times for each outer loop iteration.

Interview Connect

Understanding this algorithm shows you can use data structures smartly to improve performance from a naive O(n*k) to an optimal O(n).

This skill is valuable for solving many real-world problems efficiently.

Self-Check

What if we changed the deque to a simple list and removed elements from the front by shifting? How would the time complexity change?

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