Sliding Window Maximum Using Deque in DSA C# - Build from Scratch

Sliding Window Maximum Using Deque

📖 Scenario: You are analyzing temperature readings collected every minute. To understand the highest temperature in recent minutes, you want to find the maximum temperature in every sliding window of 3 minutes.

🎯 Goal: Build a program that uses a deque (double-ended queue) to find the maximum value in every sliding window of size 3 from a list of temperature readings.

📋 What You'll Learn

Create an integer array called temperatures with the exact values: 2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56

Create an integer variable called windowSize and set it to 3

Use a Deque<int> called deque to store indices of elements

Implement the sliding window maximum logic using the deque

Print the maximum values for each sliding window separated by spaces

💡 Why This Matters

🌍 Real World

Sliding window maximum is used in real-time data analysis like temperature monitoring, stock price analysis, and network traffic monitoring to quickly find maximum values in recent data.

💼 Career

Understanding sliding window algorithms and deque usage is important for software engineers working on performance-critical applications, real-time systems, and data stream processing.

1

Create the temperature data array

Create an integer array called temperatures with these exact values: 2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56

DSA C#

# Create the temperatures array with the given values
// Your code here

Hint

Use int[] temperatures = { ... }; with the exact numbers inside the braces.

2

Set the sliding window size

Create an integer variable called windowSize and set it to 3

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int[] temperatures = {2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56};
        // Create windowSize variable and set to 3
        // Your code here
    }
}

Hint

Use int windowSize = 3; to set the window size.

3

Implement sliding window maximum logic using deque

Create a Deque<int> called deque and write a for loop with variable i from 0 to temperatures.Length. Inside the loop, remove indices from the back of deque while the current temperature is greater than or equal to the temperature at those indices. Remove indices from the front if they are out of the current window. Add the current index i to the back of deque. When i is greater than or equal to windowSize - 1, add the temperature at the front index of deque to a list called maxValues.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int[] temperatures = {2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56};
        int windowSize = 3;
        // Implement sliding window maximum using deque
        // Your code here
    }
}

Hint

Use a Deque<int> to store indices. Remove smaller elements from the back, remove out-of-window indices from the front, add current index, and record max when window is full.

4

Print the maximum values for each sliding window

Print the values in the maxValues list separated by spaces using Console.WriteLine.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int[] temperatures = {2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56};
        int windowSize = 3;
        Deque deque = new Deque();
        List maxValues = new List();

        for (int i = 0; i  0 && temperatures[i] >= temperatures[deque.Back])
            {
                deque.RemoveFromBack();
            }

            while (deque.Count > 0 && deque.Front = windowSize - 1)
            {
                maxValues.Add(temperatures[deque.Front]);
            }
        }

        // Print the maxValues separated by spaces
        // Your code here
    }
}

public class Deque
{
    private LinkedList list = new LinkedList();

    public int Count => list.Count;

    public T Front => list.First.Value;

    public T Back => list.Last.Value;

    public void AddToBack(T item)
    {
        list.AddLast(item);
    }

    public void RemoveFromBack()
    {
        list.RemoveLast();
    }

    public void RemoveFromFront()
    {
        list.RemoveFirst();
    }
}

Hint

Use Console.WriteLine(string.Join(" ", maxValues)); to print all max values separated by spaces.

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