Sliding Window on Arrays in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the sliding window technique in arrays?

It is a method to create a 'window' of elements that moves over the array to solve problems efficiently by avoiding repeated work.

intermediate

How does the sliding window technique improve performance compared to a nested loop?

It reduces time complexity by updating the window incrementally instead of recalculating sums or values for each new window from scratch.

beginner

In a fixed-size sliding window, what happens when the window moves one step forward?

The element leaving the window is removed from the calculation, and the new element entering the window is added.

intermediate

What is a common use case for a variable-size sliding window?

Finding the smallest or largest subarray that meets a condition, like the smallest subarray with a sum greater than or equal to a target.

beginner

Why is the sliding window technique often preferred over brute force in array problems?

Because it avoids repeated calculations by reusing previous results, making the solution faster and more efficient.

What does the sliding window technique help to avoid in array problems?

A Using extra memory for a new array

B Sorting the array

C Recursion

D Repeated calculations

Sliding window avoids repeated calculations by updating results incrementally.

In a fixed-size sliding window of size 3, how many elements change when the window moves one step?

A 3

B 2

C 1

D 0

Only one element leaves and one new element enters, so effectively one element changes.

Which problem is best solved by a variable-size sliding window?

A Find smallest subarray with sum >= target

B Find max element in fixed window

C Sort the array

D Reverse the array

Variable-size windows adjust size to meet conditions like sum thresholds.

What is the main advantage of sliding window over brute force?

A Uses less memory

B Runs faster by reusing calculations

C Uses recursion

D Sorts the array first

Sliding window runs faster by updating results incrementally instead of recalculating.

Which of these is NOT a typical step in sliding window technique?

A Recalculate entire window sum every time

B Shrink the window

C Expand the window

D Move the window forward

Recalculating entire window sum every time is what sliding window avoids.

Explain how the sliding window technique works on arrays with an example.

Think about how you slide a window over a list of numbers and update sums.

Describe the difference between fixed-size and variable-size sliding windows and when to use each.

Consider problems like max sum in fixed window vs smallest subarray with sum condition.

Practice

1. What is the main purpose of the sliding window technique when working with arrays?

easy

A. To find the maximum element in the entire array

B. To sort the array elements in ascending order

C. To reverse the array elements in place

D. To efficiently process subarrays of fixed size without recomputing everything

Solution

1. Step 1: Understand the sliding window concept

   The sliding window technique moves a fixed-size window over the array to process subarrays efficiently.

2. Step 2: Compare with other options

   Sorting, reversing, or finding max in the entire array are different operations not related to sliding window.

3. Final Answer:

   To efficiently process subarrays of fixed size without recomputing everything -> Option D

4. Quick Check:

   Sliding window = efficient fixed-size subarray processing [OK]

Hint: Sliding window moves fixed-size subarray to save work [OK]

Common Mistakes:

• Confusing sliding window with sorting or reversing
• Thinking it processes the whole array at once
• Assuming it finds max without window size

2. Which of the following is the correct syntax to initialize a sliding window sum for the first k elements in JavaScript?

easy

A. let sum = arr.slice(0, k).reduce((a, b) => a + b, 0);

B. let sum = arr.reduce((a, b) => a + b, 0);

C. let sum = arr.slice(k).reduce((a, b) => a + b, 0);

D. let sum = arr.splice(0, k).reduce((a, b) => a + b, 0);

Solution

1. Step 1: Understand how to get first k elements

   Using arr.slice(0, k) extracts the first k elements without modifying the array.

2. Step 2: Sum the elements correctly

   Using reduce((a, b) => a + b, 0) sums all elements in the sliced array.

3. Final Answer:

   let sum = arr.slice(0, k).reduce((a, b) => a + b, 0); -> Option A

4. Quick Check:

   slice(0,k) + reduce sum = correct initialization [OK]

Hint: Use slice(0,k) to get first window, then reduce to sum [OK]

Common Mistakes:

• Using reduce on whole array instead of first k elements
• Using splice which modifies the original array
• Slicing from k to end instead of 0 to k

3. What will be the output of the following code?

const arr = [1, 3, 2, 6, -1, 4, 1, 8, 2];
const k = 5;
let maxSum = 0;
let windowSum = 0;

// Initialize first window sum
for(let i = 0; i < k; i++) {
  windowSum += arr[i];
}
maxSum = windowSum;

// Slide the window
for(let i = k; i < arr.length; i++) {
  windowSum += arr[i] - arr[i - k];
  if(windowSum > maxSum) maxSum = windowSum;
}
console.log(maxSum);

medium

A. 18

B. 17

C. 16

D. 19

Solution

1. Step 1: Calculate initial window sum for first 5 elements

   Sum of [1,3,2,6,-1] = 1+3+2+6-1 = 11

2. Step 2: Slide window and update sums

   Next windows and sums:
   Remove 1, add 4: 11 - 1 + 4 = 14
   Remove 3, add 1: 14 - 3 + 1 = 12
   Remove 2, add 8: 12 - 2 + 8 = 18 (maxSum updated to 18)
   Remove 6, add 2: 18 - 6 + 2 = 14

3. Step 3: Find maxSum

   Max sum found is 18

4. Final Answer:

   18 -> Option A

5. Quick Check:

   Max sliding window sum = 18 [OK]

Hint: Update sum by subtracting left, adding right element [OK]

Common Mistakes:

• Not updating maxSum correctly
• Adding instead of subtracting old element
• Wrong initial sum calculation

4. Identify the bug in the following sliding window code that finds the maximum sum of subarrays of size k:

function maxSlidingWindowSum(arr, k) {
  let maxSum = 0;
  let windowSum = 0;
  for(let i = 0; i <= k; i++) {
    windowSum += arr[i];
  }
  maxSum = windowSum;
  for(let i = k; i < arr.length; i++) {
    windowSum += arr[i] - arr[i - k];
    if(windowSum > maxSum) maxSum = windowSum;
  }
  return maxSum;
}

medium

A. The subtraction in the second loop is incorrect

B. The first loop should run while i < k, not i <= k

C. maxSum should be initialized to Number.MIN_VALUE

D. The function does not handle empty arrays

Solution

1. Step 1: Check the first loop range

   The first loop runs from i = 0 to i <= k, which means it accesses arr[k], an out-of-bound index for window size k.

2. Step 2: Correct the loop condition

   The loop should run while i < k to sum exactly k elements starting from index 0.

3. Final Answer:

   The first loop should run while i < k, not i <= k -> Option B

4. Quick Check:

   Loop range for first window = i < k [OK]

Hint: Loop for first window sums exactly k elements: i < k [OK]

Common Mistakes:

• Using <= k causes out-of-bounds error
• Ignoring array length checks
• Incorrect initialization of maxSum

5. Given an array of integers and a number k, how would you find the length of the longest subarray with sum less than or equal to k using the sliding window technique?

hard

A. Calculate prefix sums and find max subarray sum using Kadane's algorithm

B. Sort the array and use binary search to find subarrays with sum ≤ k

C. Expand the window by moving right pointer and shrink from left when sum exceeds k, track max length

D. Use nested loops to check all subarrays and pick the longest with sum ≤ k

Solution

1. Step 1: Use two pointers to create a sliding window

   Start with both pointers at the beginning, expand right pointer to add elements to sum.

2. Step 2: Shrink window from left when sum exceeds k

   If sum > k, move left pointer forward to reduce sum until sum ≤ k again, tracking max window length.

3. Step 3: Track the maximum length of valid windows

   Keep updating max length whenever sum ≤ k during expansion and shrinking.

4. Final Answer:

   Expand the window by moving right pointer and shrink from left when sum exceeds k, track max length -> Option C

5. Quick Check:

   Sliding window with two pointers and sum check = Expand the window by moving right pointer and shrink from left when sum exceeds k, track max length [OK]

Hint: Move right to expand, left to shrink when sum > k [OK]

Common Mistakes:

• Using sorting or binary search which doesn't apply here
• Using nested loops causing O(n²) time
• Confusing Kadane's algorithm with sum limit problem