Sliding Window Maximum Using Deque in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sliding Window Maximum for a Simple Array

What is the output of the following Python code that finds the maximum in each sliding window of size 3 using a deque?

DSA Python

from collections import deque

def sliding_window_max(nums, k):
    dq = deque()
    result = []
    for i in range(len(nums)):
        while dq and dq[0] <= i - k:
            dq.popleft()
        while dq and nums[dq[-1]] < nums[i]:
            dq.pop()
        dq.append(i)
        if i >= k - 1:
            result.append(nums[dq[0]])
    return result

print(sliding_window_max([1,3,-1,-3,5,3,6,7], 3))

A [3, 3, 5, 5, 6, 7]

B [1, 3, -1, -3, 5, 3]

C [3, -1, -3, 5, 3, 6]

D [7, 6, 5, 3, -3, -1]

💡 Hint

Remember the deque stores indices of elements in decreasing order of their values.

The code maintains a deque of indices where the values are in decreasing order. For each window, the front of the deque is the index of the maximum element. The output is the list of these maximums for each sliding window.

🧠 Conceptual

intermediate

Why Use a Deque for Sliding Window Maximum?

Why is a deque the best data structure to find the maximum in a sliding window efficiently?

A Because it uses hashing to find the maximum element instantly.

B Because it sorts the elements automatically, so the maximum is always at the front.

C Because it stores elements in a tree structure for quick maximum lookup.

D Because it allows adding and removing elements from both ends in O(1) time, maintaining candidates for maximum efficiently.

💡 Hint

Think about how the sliding window moves and how elements enter and leave the window.

A deque supports adding and removing elements from both ends in constant time. This allows us to keep only useful elements (potential maximums) in the deque and discard others efficiently as the window slides.

🔧 Debug

advanced

Identify the Error in Sliding Window Maximum Implementation

What error does the following code produce when run, and why? from collections import deque def sliding_window_max(nums, k): dq = deque() result = [] for i in range(len(nums)): while dq and nums[dq[-1]] <= nums[i]: dq.pop() dq.append(i) if i >= k - 1: if dq[0] <= i - k: dq.popleft() result.append(nums[dq[0]]) return result print(sliding_window_max([1,3,-1,-3,5,3,6,7], 3))

A IndexError because dq[0] is accessed before checking if dq is empty.

B TypeError because nums[dq[-1]] is compared with nums[i] incorrectly.

C Wrong output because the window removal condition is checked after appending the max.

D No error, outputs correct maximums.

💡 Hint

Check the order of removing indices outside the window and appending the maximum.

The code appends the maximum before removing indices that are out of the current window, causing the maximum to sometimes be from an old window. The removal of indices outside the window should happen before appending the maximum.

❓ Predict Output

advanced

Output of Sliding Window Maximum with Window Size 1

What is the output of this code when the window size is 1?

DSA Python

from collections import deque

def sliding_window_max(nums, k):
    dq = deque()
    result = []
    for i in range(len(nums)):
        while dq and dq[0] <= i - k:
            dq.popleft()
        while dq and nums[dq[-1]] < nums[i]:
            dq.pop()
        dq.append(i)
        if i >= k - 1:
            result.append(nums[dq[0]])
    return result

print(sliding_window_max([4, 2, 12, 3, 8], 1))

A [12]

B [4, 2, 12, 3, 8]

C [4]

D []

💡 Hint

When the window size is 1, each element is its own window.

With window size 1, the maximum in each window is the element itself, so the output is the original list.

🧠 Conceptual

expert

Time Complexity of Sliding Window Maximum Using Deque

What is the time complexity of finding the sliding window maximum using a deque for an array of length n and window size k, and why?

A O(n) because each element is added and removed from the deque at most once.

B O(n*k) because for each element, we scan k elements in the window.

C O(k*log n) because the deque maintains a sorted order using binary search.

D O(n^2) because nested loops are used to find maximums.

💡 Hint

Consider how many times each element can be pushed or popped from the deque.

Each element is pushed into the deque once and popped at most once, so the total operations are proportional to n, making the algorithm O(n).