Sliding Window on Arrays in DSA Javascript - Time & Space Complexity

Time Complexity: Sliding Window on Arrays

O(n)

Understanding Time Complexity

We want to understand how fast the sliding window technique works when processing arrays.

Specifically, how the number of steps grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxSumSubarray(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;
}
    

This code finds the maximum sum of any subarray of size k using a sliding window.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops that each run over parts of the array.
• How many times: First loop runs k times, second loop runs (n - k) times, where n is array length.

How Execution Grows With Input

As the array size grows, the total steps grow roughly in a straight line with n.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase directly with input size, not faster or slower.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line as the array gets bigger.

Common Mistake

[X] Wrong: "The nested loop inside the sliding window makes this O(n²)."

[OK] Correct: The code only uses two separate loops, not nested loops, so it does not multiply steps but adds them, keeping it linear.

Interview Connect

Knowing how sliding window runs efficiently helps you solve many array problems quickly and confidently in interviews.

Self-Check

"What if we changed the window size k to be equal to the array length n? How would the time complexity change?"

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