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

How does the algorithm handle many elements efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void slidingWindowMax(int arr[], int n, int k) {
    int deque[n];
    int front = 0, back = -1;
    for (int i = 0; i < n; i++) {
        while (front <= back && arr[deque[back]] < arr[i]) back--;
        deque[++back] = i;
        if (deque[front] == i - k) front++;
        if (i >= k - 1) printf("%d ", arr[deque[front]]);
    }
}
    

This code finds the maximum value in every window of size k in the array using a deque.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	About 20 operations
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 complete grows directly in proportion to the number of elements.

Common Mistake

[X] Wrong: "Since there is a nested while 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 inner loop does not run n times for each element.

Interview Connect

Understanding this efficient approach shows you can handle problems that need fast sliding window calculations, a common real-world need.

Self-Check

"What if we used a simple array instead of a deque to track maximums? 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 store the sum of elements in each window

B. To sort the array elements before processing

C. To reverse the array for easier traversal

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

Solution

1. Step 1: Understand the role of deque

   A deque stores indices of elements in a way that the front always has the index of the maximum element for the current window.

2. Step 2: Why deque helps in O(1) max retrieval

   Because the deque maintains elements in decreasing order, the maximum is always at the front, allowing quick access without scanning the window.

3. Final Answer:

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

4. Quick Check:

   Deque stores max indices for O(1) max retrieval [OK]

Hint: Deque keeps max at front for quick max retrieval [OK]

Common Mistakes:

• Thinking deque stores actual values instead of indices
• Using deque to sort elements
• Using deque to store sums

2. Which of the following is the correct way to declare a deque in C for storing integer indices?

easy

A. struct deque dq; // assuming a deque struct is defined

B. deque<int> dq;

C. int dq[];

D. deque dq = new deque();

Solution

1. Step 1: Understand C data structure declaration

   C does not have built-in deque type; we use structs or arrays to implement it.

2. Step 2: Identify correct syntax

   struct deque dq; // assuming a deque struct is defined shows a struct declaration which is the correct way in C if a deque struct is defined.

3. Final Answer:

   struct deque dq; // assuming a deque struct is defined -> Option A

4. Quick Check:

   C uses structs for custom data types [OK]

Hint: Use struct for deque in C, no built-in deque type [OK]

Common Mistakes:

• Using C++ syntax like deque<int>
• Trying to declare deque like an array without size
• Using Java or other language syntax

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

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

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

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

Solution

1. Step 1: Slide window and find max for each window of size 4

   Windows and max values:
   Window 1: [2,1,3,4] max=4
   Window 2: [1,3,4,6] max=6
   Window 3: [3,4,6,3] max=6
   Window 4: [4,6,3,8] max=8
   Window 5: [6,3,8,9] max=9
   Window 6: [3,8,9,10] max=10
   Window 7: [8,9,10,12] max=12
   Window 8: [9,10,12,56] max=56

2. Step 2: Collect all max values in order

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

3. Final Answer:

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

4. Quick Check:

   Max per window = [4,6,6,8,9,10,12,56] [OK]

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

Common Mistakes:

• Including partial windows smaller than size 4
• Off-by-one errors in window indexing
• Confusing max values with sums

4. Given this snippet for sliding window maximum, what is the bug?

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)
    printf("%d ", arr[dq.front()]);
}

medium

A. The condition to pop front should be < instead of <=

B. No bug, code is correct

C. The condition to pop front should be checked before pushing current index

D. The order of popping from back and pushing is wrong

Solution

1. Step 1: Analyze deque maintenance logic

   The code pops smaller elements from back before pushing current index, ensuring decreasing order.

2. Step 2: Check window boundary condition

   The code pops front if the front index is out of current window (<= i-k), which is correct.

3. Step 3: Confirm output printing

   Printing starts after first full window (i >= k-1), printing max at front index.

4. Final Answer:

   No bug, code is correct -> Option B

5. Quick Check:

   Deque logic and conditions are correct [OK]

Hint: Check deque front removal after pushing current index [OK]

Common Mistakes:

• Removing front before pushing current index
• Using wrong comparison operators in conditions
• Printing output before full window

5. How would you modify the sliding window maximum algorithm using deque to find the sliding window minimum instead?

hard

A. Change the comparison from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] when popping from back

B. Reverse the array before applying the same algorithm

C. Use a stack instead of a deque

D. Change the window size to half

Solution

1. Step 1: Understand max vs min logic in deque

   For max, we pop from back while elements are smaller or equal. For min, we pop while elements are greater or equal.

2. Step 2: Modify comparison operator

   Change condition from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] to maintain increasing order for min.

3. Final Answer:

   Change the comparison from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] when popping from back -> Option A

4. Quick Check:

   Reverse comparison for min sliding window [OK]

Hint: Flip comparison sign to find min instead of max [OK]

Common Mistakes:

• Reversing array instead of changing comparison
• Using stack which doesn't maintain order
• Changing window size unnecessarily