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

Time Complexity: Sliding Window Maximum Using Deque

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum in each sliding window changes as the input size grows.

Specifically, how fast can we find maximums when moving a window over an array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

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

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over all elements in nums.
• How many times: Each element is added and removed from the deque at most once.

How Execution Grows With Input

As the input size n grows, each element is processed a limited number of times, so the total work grows roughly in direct proportion to n.

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

Pattern observation: The operations grow linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to find all sliding window maximums grows directly with the size of the input list.

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 pushed and popped from the deque only once, so the total number of operations inside the while loop across the entire run is at most n, not n times n.

Interview Connect

Understanding this efficient approach shows you can optimize problems by cleverly managing data, a skill highly valued in real-world coding challenges.

Self-Check

What if we replaced the deque with a simple list and searched for the maximum in each window? How would the time complexity change?

Practice

1. What is the main purpose of using a deque in the Sliding Window Maximum problem?

easy

A. To reverse the array elements

B. To sort the entire array before processing

C. To store the sum of elements in the window

D. To store indices of elements in a way that helps find the maximum efficiently

Solution

1. Step 1: Understand the role of deque

   A deque stores indices of elements in the current window, maintaining order to quickly find the max.

2. Step 2: Why not other options?

   Sorting or reversing is not done; sum is unrelated to max. Deque helps track max efficiently.

3. Final Answer:

   To store indices of elements in a way that helps find the maximum efficiently -> Option D

4. Quick Check:

   Deque stores indices for max tracking [OK]

Hint: Deque keeps max candidates in order for quick max retrieval [OK]

Common Mistakes:

• Thinking deque stores actual values instead of indices
• Assuming deque sorts the array
• Confusing deque with sum calculation

2. Which of the following is the correct syntax to initialize an empty deque in Python for the sliding window maximum problem?

easy

A. from collections import deque; dq = deque()

B. deque = []

C. dq = deque([])

D. dq = collections.deque()

Solution

1. Step 1: Import deque correctly

   Deque must be imported from collections module before use.

2. Step 2: Initialize empty deque

   Using dq = deque() after import is correct syntax; dq = deque([]) also works but less common.

3. Final Answer:

   from collections import deque; dq = deque() -> Option A

4. Quick Check:

   Import then dq = deque() [OK]

Hint: Always import deque from collections before initializing [OK]

Common Mistakes:

• Using list [] instead of deque
• Not importing deque before use
• Using collections.deque() without import

3. What is the output of the following code snippet?

from collections import deque

def max_sliding_window(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(max_sliding_window([1,3,-1,-3,5,3,6,7], 3))

medium

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

B. [1, 3, -1, -3, 5, 3, 6, 7]

C. [3, -1, -3, 5, 3, 6, 7]

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

Solution

1. Step 1: Understand sliding window max logic

   For each window of size 3, the code finds max by maintaining indices in dq.

2. Step 2: Trace output for input

   Windows and max: [1,3,-1] -> 3, [3,-1,-3] -> 3, [-1,-3,5] -> 5, [-3,5,3] -> 5, [5,3,6] -> 6, [3,6,7] -> 7

3. Final Answer:

   [3, 3, 5, 5, 6, 7] -> Option A

4. Quick Check:

   Sliding window max = [3,3,5,5,6,7] [OK]

Hint: Track max in each window by deque front index [OK]

Common Mistakes:

• Appending values instead of indices to deque
• Not removing indices out of window
• Confusing window size with output length

4. Identify the bug in this sliding window maximum code snippet:

from collections import deque

def max_sliding_window(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 dq[0] == i - k:
            dq.popleft()
        if i >= k - 1:
            result.append(nums[dq[0]])
    return result

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

medium

A. The condition to remove old indices should be dq[0] < i - k

B. The condition to remove old indices should be dq[0] <= i - k

C. The condition to remove old indices should be dq[0] == i - k + 1

D. The condition to remove old indices should be dq[0] <= i - k + 1

Solution

1. Step 1: Check condition for removing old indices

   Indices outside the window must be removed when dq[0] <= i - k (equivalent to dq[0] < i - k + 1).

2. Step 2: Analyze given condition

   Given condition dq[0] == i - k removes only if exactly equal, missing older indices or multiples; must use while loop too.

3. Final Answer:

   The condition to remove old indices should be dq[0] <= i - k -> Option B

4. Quick Check:

   Remove indices outside window with dq[0] <= i - k [OK]

Hint: Remove indices strictly less than window start index [OK]

Common Mistakes:

• Using equality instead of less than for removal
• Removing indices too late causing wrong max
• Not removing old indices at all

5. Given an array nums = [4, 2, 12, 3, 8, 7, 9] and window size k = 4, what is the output of the sliding window maximum using deque approach?

hard

A. [4, 12, 8, 9]

B. [4, 12, 7, 9]

C. [12, 12, 12, 9]

D. [12, 12, 8, 9]

Solution

1. Step 1: Identify windows and their max values

   Windows of size 4: [4,2,12,3] max=12, [2,12,3,8] max=12, [12,3,8,7] max=12, [3,8,7,9] max=9.

2. Step 2: Verify max values carefully

   Third window max is 12, but 12 is at index 2, window covers indices 2-5: 12,3,8,7 max=12. Fourth window max is 9.

3. Final Answer:

   [12, 12, 12, 9] -> Option C

4. Quick Check:

   Sliding window max for k=4 = [12,12,12,9] [OK]

Hint: Check max in each window carefully by indices [OK]

Common Mistakes:

• Mixing up window boundaries
• Choosing wrong max by ignoring window size
• Confusing values with indices