Why Sliding Window Maximum Using Deque in DSA C#?

The Big Idea

Discover how a simple data structure can save you from endless repeated work!

The Scenario

Imagine you have a long list of daily temperatures, and you want to find the hottest day in every 3-day period. Doing this by checking each group of 3 days one by one feels like a lot of repeated work.

The Problem

Manually checking each 3-day window means looking at many overlapping days again and again. This wastes time and can cause mistakes, especially if the list is very long.

The Solution

Using a deque (a special list where you can add or remove items from both ends) helps keep track of the hottest days efficiently. It remembers only the important days, so you don't have to check all days every time.

Before vs After

✗ Before

for (int i = 0; i <= nums.Length - k; i++) {
    int max = nums[i];
    for (int j = i; j < i + k; j++) {
        if (nums[j] > max) max = nums[j];
    }
    Console.Write(max + " ");
}

✓ After

LinkedList<int> deque = new LinkedList<int>();
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)
        Console.Write(nums[deque.First.Value] + " ");
}

What It Enables

This method lets you find the maximum in every sliding window quickly, even for very large lists, saving time and effort.

Real Life Example

In stock trading, you might want to know the highest price of a stock in the last 10 minutes at every minute. Using this method helps you get that instantly without checking all prices every time.

Key Takeaways

Manual checking repeats work and is slow.

Deque keeps track of important values efficiently.

Sliding window maximum becomes fast and easy.

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