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

Time Complexity: Sliding Window Maximum Using Deque

O(n)

Understanding Time Complexity

We want to understand how fast we can find the maximum value in every sliding window of a list.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <deque>
#include <vector>
using namespace std;

vector<int> maxSlidingWindow(vector<int>& nums, int k) {
    deque<int> dq;
    vector<int> result;
    for (int i = 0; i < nums.size(); i++) {
        while (!dq.empty() && dq.front() <= i - k) dq.pop_front();
        while (!dq.empty() && nums[dq.back()] < nums[i]) dq.pop_back();
        dq.push_back(i);
        if (i >= k - 1) result.push_back(nums[dq.front()]);
    }
    return result;
}
    

This code finds the maximum value in each sliding window of size k using a double-ended queue.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over the input array from start to end.
• How many times: Exactly once for each element (n times), with some extra pops from the deque inside the loop.

How Execution Grows With Input

Each element is added and removed from the deque at most once, so the total operations grow roughly in a straight line with input size.

Input Size (n)	Approx. Operations
10	About 20 operations (adding and removing elements)
100	About 200 operations
1000	About 2000 operations

Pattern observation: The operations increase roughly in direct proportion to the input size.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Because there are nested while loops, the time complexity must be O(n²)."

[OK] Correct: Each element is pushed and popped at most once, so the total work inside the loops is still proportional to n, not n squared.

Interview Connect

Understanding this technique shows you can optimize problems by cleverly using data structures to avoid repeated work.

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

easy

A. To store indices of elements in a way that helps find the maximum in O(1) time per window

B. To sort the entire array before processing windows

C. To reverse the array for easier maximum calculation

D. To store the actual maximum values of each window directly

Solution

1. Step 1: Understand the role of deque

   The deque stores indices of elements in the current window, maintaining them in decreasing order of their values.

2. Step 2: How deque helps find maximum

   The front of the deque always holds the index of the maximum element for the current window, allowing O(1) access.

3. Final Answer:

   To store indices of elements in a way that helps find the maximum in O(1) time per window -> Option A

4. Quick Check:

   Deque stores indices for max retrieval [OK]

Hint: Deque keeps max candidates' indices for quick max retrieval [OK]

Common Mistakes:

• Thinking deque stores actual values instead of indices
• Assuming deque sorts the entire array
• Believing deque reverses the array

2. Which of the following C++ deque operations is used to remove elements from the back while maintaining the sliding window maximum?

easy

A. push_front()

B. pop_front()

C. pop_back()

D. push_back()

Solution

1. Step 1: Identify deque operation for removing from back

   pop_back() removes the last element from the deque, which is used to discard smaller elements behind the current element.

2. Step 2: Why pop_back() is used here

   We remove elements from the back that are smaller than the current element to maintain decreasing order in the deque.

3. Final Answer:

   pop_back() -> Option C

4. Quick Check:

   pop_back() removes smaller elements from back [OK]

Hint: Use pop_back() to remove smaller elements behind current [OK]

Common Mistakes:

• Using pop_front() instead of pop_back() to remove from back
• Confusing push operations with pop operations
• Removing elements from front incorrectly

3. What is the output of the sliding window maximum for array [2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56] with window size 4?

medium

A. [2, 3, 4, 6, 8, 9, 10, 12, 56]

B. [3, 4, 6, 6, 8, 9, 10, 12, 56]

C. [3, 4, 6, 8, 9, 10, 12]

D. [4, 6, 6, 8, 9, 10, 12, 56]

Solution

1. Step 1: Slide window and find max for each window

   Windows and max values: [2,1,3,4] -> 4, [1,3,4,6] -> 6, [3,4,6,3] -> 6, [4,6,3,8] -> 8, [6,3,8,9] -> 9, [3,8,9,10] -> 10, [8,9,10,12] -> 12, [9,10,12,56] -> 56

2. Step 2: Collect all max values in order

   The max values for each window are [4, 6, 6, 8, 9, 10, 12, 56]

3. Final Answer:

   [4, 6, 6, 8, 9, 10, 12, 56] -> Option D

4. Quick Check:

   Sliding window max matches [4, 6, 6, 8, 9, 10, 12, 56] [OK]

Hint: Slide window, track max with deque front [OK]

Common Mistakes:

• Confusing window size or missing last windows
• Incorrectly calculating max for each window
• Off-by-one errors in output length

4. Identify the bug in this snippet for sliding window maximum using deque:

for (int i = 0; i < n; i++) {
  while (!dq.empty() && arr[dq.back()] <= arr[i]) dq.pop_back();
  dq.push_back(i);
  if (dq.front() == i - k) dq.pop_front();
  if (i >= k - 1) result.push_back(arr[dq.front()]);
}

medium

A. The condition to pop front should be dq.front() <= i - k - 1

B. The condition to pop front should be dq.front() < i - k

C. The condition to pop front should be dq.front() == i - k + 1

D. The condition to pop front should be dq.front() < i - k + 1

Solution

1. Step 1: Understand when to remove front index

   The front index should be removed if it is out of the current window. The window covers indices [i-k+1, i]. So indices less than i-k+1 are out.

2. Step 2: Correct condition for popping front

   The condition should be dq.front() <= i - k to remove indices outside the window.

3. Final Answer:

   The condition to pop front should be dq.front() <= i - k -> Option B

4. Quick Check:

   Pop front if index outside window [OK]

Hint: Pop front if index <= window start - 1 (i-k) [OK]

Common Mistakes:

• Using equality instead of less than for popping front
• Off-by-one errors in window boundary
• Not removing outdated indices from deque

5. Given an array [1, 3, -1, -3, 5, 3, 6, 7] and window size 3, how would you modify the deque approach to also return the minimum in each sliding window along with the maximum?

hard

A. Use two deques: one for max (decreasing) and one for min (increasing) values

B. Use one deque and store pairs of (max, min) values

C. Sort each window to find max and min

D. Use a priority queue for max and a stack for min

Solution

1. Step 1: Understand separate tracking for max and min

   To get both max and min efficiently, maintain two deques: one keeps indices of max candidates in decreasing order, the other keeps indices of min candidates in increasing order.

2. Step 2: How two deques work together

   For each window, the front of max deque gives max, front of min deque gives min. Both deques are updated similarly but with opposite ordering.

3. Final Answer:

   Use two deques: one for max (decreasing) and one for min (increasing) values -> Option A

4. Quick Check:

   Two deques track max and min separately [OK]

Hint: Maintain two deques: max decreasing, min increasing [OK]

Common Mistakes:

• Trying to store both max and min in one deque
• Sorting each window (inefficient)
• Using wrong data structures for min tracking