Sliding Window Maximum Using Deque in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

A deque helps keep track of indices of elements in the current window in a way that the front always holds the index of the maximum element. It allows efficient insertion and removal from both ends to maintain the window's maximum.

beginner

In the Sliding Window Maximum problem, why do we remove elements from the back of the deque when they are smaller than the current element?

Because smaller elements cannot be the maximum if a bigger element comes after them in the window. Removing them keeps the deque elements in decreasing order, ensuring the front is always the maximum.

beginner

What does the deque store in the Sliding Window Maximum algorithm?

The deque stores indices of elements from the array, not the elements themselves. This helps track which elements are inside the current window and their positions.

intermediate

How do we ensure the deque only contains elements from the current sliding window?

We remove indices from the front of the deque if they are outside the current window range (i.e., their index is less than the current index minus window size plus one).

intermediate

What is the time complexity of the Sliding Window Maximum algorithm using a deque and why?

The time complexity is O(n) because each element is added and removed from the deque at most once, making the operations linear in total.

What does the front of the deque represent in the Sliding Window Maximum algorithm?

A The last element added to the window

B Index of the minimum element in the current window

C Index of the maximum element in the current window

D The size of the current window

The front of the deque always holds the index of the maximum element in the current sliding window.

Why do we remove elements from the back of the deque when they are smaller than the current element?

A Because smaller elements cannot be maximum if a bigger element is after them

B To maintain the order of elements as they appear in the array

C To reduce the size of the deque for memory optimization

D To keep the deque sorted in increasing order

Removing smaller elements from the back ensures the deque stores elements in decreasing order, so the maximum is always at the front.

What do we do when the index at the front of the deque is outside the current window?

A Remove it from the back

B Remove it from the front

C Keep it in the deque

D Replace it with the current index

We remove indices from the front if they are outside the current window to keep the deque valid.

What is stored inside the deque in the Sliding Window Maximum algorithm?

A The actual values of the elements

B The minimum values only

C The maximum values only

D Indices of elements

The deque stores indices to track positions and window boundaries efficiently.

What is the overall time complexity of the Sliding Window Maximum algorithm using a deque?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

Each element is added and removed at most once, so the algorithm runs in linear time O(n).

Explain how a deque helps solve the Sliding Window Maximum problem efficiently.

Think about how the deque keeps track of the maximum without checking all elements every time.

Describe the step-by-step process of updating the deque when moving the sliding window by one position.

Focus on what happens to the deque when the window slides forward.

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

B. To sort the entire array before processing

C. To reverse the array elements

D. To store the sum of elements in the window

Solution

1. Step 1: Understand the role of deque in sliding window

   The 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 -> Option A

4. Quick Check:

   Deque stores max indices = To store indices of elements in a way that helps find the maximum in O(1) time [OK]

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

Common Mistakes:

• Thinking deque stores actual values instead of indices
• Assuming deque sorts the entire array
• Confusing deque with a stack or queue only

2. Which of the following C# code snippets correctly initializes a deque to be used for sliding window maximum?

easy

A. var deque = new Stack<int>();

B. var deque = new LinkedList<int>();

C. var deque = new Queue<int>();

D. var deque = new List<int>();

Solution

1. Step 1: Identify the correct data structure for deque in C#

   C# does not have a built-in deque class, but LinkedList<T> supports adding/removing from both ends, acting as a deque.

2. Step 2: Why LinkedList is suitable

   LinkedList allows efficient insertion and removal from front and back, which is essential for deque operations in sliding window maximum.

3. Final Answer:

   var deque = new LinkedList<int>(); -> Option B

4. Quick Check:

   LinkedList supports deque operations = var deque = new LinkedList<int>(); [OK]

Hint: Use LinkedList for deque in C# since no built-in deque [OK]

Common Mistakes:

• Using Queue which only supports FIFO
• Using Stack which only supports LIFO
• Using List which is inefficient for front removals

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

medium

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

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

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

D. [3, 4, 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: Write the max values as output array

   The output array is [4, 6, 6, 8, 9, 10, 12, 56].

3. Final Answer:

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

4. Quick Check:

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

Hint: Check max in each window carefully, watch overlapping elements [OK]

Common Mistakes:

• Confusing max with last element in window
• Missing last window max
• Off-by-one errors in window sliding

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

for (int i = 0; i < nums.Length; i++) {
  while (deque.Count > 0 && nums[deque.Last.Value] <= nums[i]) {
    deque.RemoveLast();
  }
  deque.AddLast(i);
  if (deque.First.Value >= i - k) {
    deque.RemoveFirst();
  }
  if (i >= k - 1) {
    result.Add(nums[deque.First.Value]);
  }
}

medium

A. The deque should store values, not indices

B. The condition to remove from front should be < instead of <=

C. The condition to remove from back should be >= instead of <=

D. The condition to remove from front should be <= instead of >=

Solution

1. Step 1: Analyze condition for removing front index

   The condition 'deque.First.Value >= i - k' removes indices inside the window since valid front indices are >= i - k + 1 which is always true for >= i - k, causing incorrect removal of current max candidates.

2. Step 2: Correct condition to remove outdated indices

   Window covers [i-k+1 to i], remove if deque.First.Value < i - k + 1 i.e. deque.First.Value <= i - k.

3. Final Answer:

   The condition to remove from front should be <= instead of >= -> Option D

4. Quick Check:

   Remove outdated indices with <= i - k condition [OK]

Hint: Check window boundary condition carefully for removing front index [OK]

Common Mistakes:

• Using >= instead of <= causes removing valid indices
• Removing from back incorrectly
• Storing values instead of indices in deque

5. Given an array nums = [1,3,-1,-3,5,3,6,7] and window size k = 3, how would you modify the sliding window maximum algorithm using deque to also return the indices of the maximum elements for each window?

hard

A. Store values in deque and output their positions from the array

B. Store indices in deque and output nums[deque.Last.Value] for each window

C. Store indices in deque and output deque.First.Value for each window

D. Store values in deque and output the maximum value only

Solution

1. Step 1: Understand what deque stores

   The deque stores indices of elements, not values, to track positions and efficiently remove outdated indices.

2. Step 2: Output indices of maximum elements

   The maximum element's index is always at the front of the deque, so outputting deque.First.Value for each window gives the index of the max element.

3. Final Answer:

   Store indices in deque and output deque.First.Value for each window -> Option C

4. Quick Check:

   Deque front index = max element index [OK]

Hint: Deque front holds max index; output it per window [OK]

Common Mistakes:

• Storing values instead of indices in deque
• Outputting deque.Last.Value instead of First
• Confusing values with indices in output