Sliding Window Maximum Using Deque in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the deque for sliding window maximum.

DSA Python

from collections import deque

dq = [1]()

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Alist

Bset

Cdeque

Ddict

Attempts:

3 left

💡 Hint

Common Mistakes

Using list instead of deque causes inefficient pops from the front.

Using set or dict is incorrect because they don't maintain order.

We use deque from collections to efficiently add and remove elements from both ends.

2fill in blank

medium

Complete the code to remove indices from the deque that are out of the current window.

DSA Python

if dq and dq[0] [1] i - k:
    dq.popleft()

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A<=

B>=

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' instead of '<=' causes wrong removal.

Not removing old indices leads to incorrect maximum.

We remove the index at the front if it is outside the current window, which means its index is less than or equal to i - k.

3fill in blank

hard

Fix the error in the loop that removes smaller elements from the deque.

DSA Python

while dq and nums[dq[-1]] [1] nums[i]:
    dq.pop()

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A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' or '>' causes incorrect removals.

Using '<=' keeps smaller elements which is wrong.

We remove indices from the deque while the current number is greater than the numbers at those indices to maintain decreasing order.

4fill in blank

hard

Fill both blanks to append the maximum for each window to the result list.

DSA Python

if i [1] k - 1:
    result.[2](nums[dq[0]])

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A>=

Bappend

C<=

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>=' delays adding results.

Using 'pop' instead of 'append' removes elements incorrectly.

We start adding maximums to the result list once the first full window is processed (i >= k - 1). We use append to add the maximum.

5fill in blank

hard

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

DSA Python

def maxSlidingWindow(nums, k):
    from collections import deque
    dq = deque()
    result = []
    for i in range(len(nums)):
        if dq and dq[0] [1] i - k:
            dq.popleft()
        while dq and nums[dq[-1]] [2] nums[i]:
            dq.pop()
        dq.append(i)
        if i [3] k - 1:
            result.append(nums[dq[0]])
    return result

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

A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing comparison operators causing wrong removals.

Appending results before the window is full.

The first blank removes indices out of the window (less than or equal to i - k), the second blank removes smaller elements (<), and the third blank checks if the window is full (i >= k - 1) to append the maximum.