Why Sliding Window Maximum Using Deque in DSA Javascript?

The Big Idea

Discover how a simple queue can speed up finding the hottest days in your data!

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 3-day group one by one feels like repeating the same work over and over.

The Problem

Manually checking each group means looking at many numbers again and again. This takes a lot of time and can easily cause mistakes, especially if the list is very long.

The Solution

Using a sliding window with a deque helps keep track of the hottest days efficiently. It remembers only the important days and quickly updates as the window moves, saving time and effort.

Before vs After

✗ Before

function maxSlidingWindow(nums, k) {
  let result = [];
  for (let i = 0; i <= nums.length - k; i++) {
    let max = Math.max(...nums.slice(i, i + k));
    result.push(max);
  }
  return result;
}

✓ After

function maxSlidingWindow(nums, k) {
  const deque = [];
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    while (deque.length && nums[deque[deque.length - 1]] <= nums[i]) deque.pop();
    deque.push(i);
    if (deque[0] <= i - k) deque.shift();
    if (i >= k - 1) result.push(nums[deque[0]]);
  }
  return result;
}

What It Enables

This method lets you find maximum values in moving windows quickly, even for very large lists.

Real Life Example

Stock traders use this to find the highest stock price in the last 10 days as new prices come in every day.

Key Takeaways

Manual checking repeats work and is slow.

Deque keeps track of important elements efficiently.

Sliding window maximum helps analyze data streams fast.

Practice

1. What is the main purpose of using a deque in the Sliding Window Maximum algorithm?

easy

A. To sort the entire array before processing

B. To store all elements of the array for quick access

C. To keep track of indices of elements in the current window in decreasing order of their values

D. To reverse the array elements

Solution

1. Step 1: Understand the role of deque in the algorithm

   The deque stores indices of elements in the current window, maintaining them in decreasing order of their values to quickly find the maximum.

2. Step 2: Why other options are incorrect

   To store all elements of the array for quick access is incorrect because deque does not store all elements, only relevant indices. To sort the entire array before processing and D do not relate to the sliding window maximum logic.

3. Final Answer:

   To keep track of indices of elements in the current window in decreasing order of their values -> Option C

4. Quick Check:

   Deque stores indices in decreasing order for max [OK]

Hint: Deque keeps max candidates in order for O(1) 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 without order

2. Which of the following is the correct syntax to initialize an empty deque in JavaScript for the Sliding Window Maximum algorithm?

easy

A. const deque = new ArrayDeque();

B. const deque = new Deque();

C. const deque = new Queue();

D. const deque = [];

Solution

1. Step 1: Identify JavaScript data structure for deque

   JavaScript does not have a built-in Deque class, so an empty array is used to simulate a deque.

2. Step 2: Why other options are invalid

   Options A, B, and D refer to classes not available in standard JavaScript.

3. Final Answer:

   const deque = []; -> Option D

4. Quick Check:

   Use array as deque in JS [OK]

Hint: Use array as deque in JavaScript [OK]

Common Mistakes:

• Trying to use non-existent Deque class
• Using Queue which lacks double-ended operations
• Confusing JavaScript with other languages' syntax

3. What is the output of the following code snippet?

function maxSlidingWindow(nums, k) {
  const deque = [];
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    while (deque.length && nums[deque[deque.length - 1]] <= nums[i]) {
      deque.pop();
    }
    deque.push(i);
    if (deque[0] <= i - k) {
      deque.shift();
    }
    if (i >= k - 1) {
      result.push(nums[deque[0]]);
    }
  }
  return result;
}

console.log(maxSlidingWindow([1,3,-1,-3,5,3,6,7], 3));

medium

A. [1, 3, -1, -3, 5, 3]

B. [3, 3, 5, 5, 6, 7]

C. [7, 6, 5, 3, -3, -1]

D. [3, 5, 6, 7]

Solution

1. Step 1: Trace the sliding window maximum for each window

   Windows of size 3 and their max values are: [1,3,-1] -> 3, [3,-1,-3] -> 3, [-1,-3,5] -> 5, [-3,5,3] -> 5, [5,3,6] -> 6, [3,6,7] -> 7.

2. Step 2: Confirm the result array matches the expected output

   The result array is [3, 3, 5, 5, 6, 7], matching [3, 3, 5, 5, 6, 7].

3. Final Answer:

   [3, 3, 5, 5, 6, 7] -> Option B

4. Quick Check:

   Sliding window max = [3,3,5,5,6,7] [OK]

Hint: Track max in each window using deque indices [OK]

Common Mistakes:

• Not removing indices outside the window
• Pushing values instead of indices into deque
• Incorrectly updating the result array

4. Identify the bug in the following Sliding Window Maximum code snippet:

function maxSlidingWindow(nums, k) {
  const deque = [];
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    while (deque.length && nums[deque[0]] <= nums[i]) {
      deque.shift();
    }
    deque.push(i);
    if (deque[0] <= i - k) {
      deque.shift();
    }
    if (i >= k - 1) {
      result.push(nums[deque[0]]);
    }
  }
  return result;
}

medium

A. The condition in the while loop should compare with the last element, not the first

B. The deque should store values, not indices

C. The if condition to remove out-of-window indices is missing

D. The result array is not being returned

Solution

1. Step 1: Analyze the while loop condition

   The while loop incorrectly compares nums[deque[0]] (front) with nums[i], but it should compare with nums[deque[deque.length - 1]] (back) to maintain decreasing order.

2. Step 2: Confirm other parts are correct

   The deque stores indices correctly, the out-of-window removal condition exists, and the result is returned properly.

3. Final Answer:

   The condition in the while loop should compare with the last element, not the first -> Option A

4. Quick Check:

   Compare with deque's last element in while loop [OK]

Hint: While loop must check deque's last element for correct order [OK]

Common Mistakes:

• Comparing with front instead of back in deque
• Storing values instead of indices in deque
• Forgetting to remove indices outside the window

5. Given an array nums = [4, 2, 12, 3, 8, 7, 9] and window size k = 4, what will be the output of the Sliding Window Maximum algorithm using a deque?

hard

A. [12, 12, 12, 9]

B. [4, 12, 8, 9]

C. [12, 3, 8, 7]

D. [4, 2, 12, 3]

Solution

1. Step 1: Identify each window and its maximum

   Windows of size 4 are: [4,2,12,3] max=12, [2,12,3,8] max=12, [12,3,8,7] max=12, [3,8,7,9] max=9.

2. Step 2: Collect the maximums in order

   The result array is [12, 12, 12, 9], matching [12, 12, 12, 9].

3. Final Answer:

   [12, 12, 12, 9] -> Option A

4. Quick Check:

   Max per window = [12,12,12,9] [OK]

Hint: Check max in each window by comparing edges with deque [OK]

Common Mistakes:

• Not sliding window correctly by one index
• Forgetting to remove old indices from deque
• Mixing values and indices in deque