Sliding Window Maximum Using Deque in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a deque to store indices.

DSA C#

var deque = new [1]<int>();

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

AList

BQueue

CStack

DLinkedList

Attempts:

3 left

💡 Hint

Common Mistakes

Using List which is not efficient for removing from front.

Using Stack or Queue which only support one end operations.

We use LinkedList<int> in C# to implement a deque because it allows adding and removing from both ends efficiently.

2fill in blank

medium

Complete the code to remove indices from the back while current element is greater.

DSA C#

while (deque.Count > 0 && nums[deque.Last.Value] [1] nums[i])
{
    deque.RemoveLast();
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < causes wrong elements to be removed.

Using >= or <= changes the strictness and may cause incorrect results.

We remove indices from the back while the current element is greater than the element at those indices to maintain decreasing order.

3fill in blank

hard

Fix the error in removing indices out of the current window from the front.

DSA C#

if (deque.Count > 0 && deque.First.Value [1]= i - k + 1)
{
    deque.RemoveFirst();
}

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

A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or < causes off-by-one errors.

Using <= removes indices too early or late.

We remove from the front if the index is less than or equal to the start of the current window (i - k + 1).

4fill in blank

hard

Fill both blanks to add the current index and assign the max to result.

DSA C#

deque.[1](i);
if (i >= k - 1)
{
    result[i - k + 1] = nums[deque.[2].Value];
}

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

AAddLast

BAddFirst

CFirst

DLast

Attempts:

3 left

💡 Hint

Common Mistakes

Adding index to front instead of back.

Taking max from the back instead of front.

We add the current index to the back of the deque and take the front index's value as the max for the current window.

5fill in blank

hard

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

DSA C#

public int[] MaxSlidingWindow(int[] nums, int k)
{
    if (nums == null || nums.Length == 0) return new int[0];
    var result = new int[nums.Length - k + 1];
    var deque = new LinkedList<int>();
    for (int i = 0; i < nums.Length; i++)
    {
        while (deque.Count > 0 && nums[deque.Last.Value] [1] nums[i])
            deque.RemoveLast();
        deque.[2](i);
        if (deque.First.Value [3] i - k)
            deque.RemoveFirst();
        if (i >= k - 1)
            result[i - k + 1] = nums[deque.First.Value];
    }
    return result;
}

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

A<

B>

CAddLast

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators causing incorrect removals.

Adding index to front instead of back.

Incorrect condition for removing out-of-window indices.

The function removes smaller elements from the back (<), adds current index to the back (AddLast), and removes indices out of window from the front (<=).

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