Sliding Window on Arrays in DSA C# - Time & Space Complexity

Time Complexity: Sliding Window on Arrays

O(n)

Understanding Time Complexity

We want to understand how the time needed changes when using the sliding window technique on arrays.

Specifically, how does the number of steps grow as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int MaxSumSubarray(int[] arr, int k) {
    int maxSum = 0, windowSum = 0;
    for (int i = 0; i < k; i++) {
        windowSum += arr[i];
    }
    maxSum = windowSum;
    for (int 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 work grows directly with the size of the array.

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 inner loop makes this O(n*k) because it looks like nested loops."

[OK] Correct: The loops are separate, not nested, so total steps add up, not multiply.

Interview Connect

Understanding sliding window time complexity helps you explain efficient solutions clearly and confidently.

Self-Check

"What if we changed the window size k to vary with n, like k = n/2? 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 efficiently find subarray sums or properties without recalculating everything each time

B. To sort the array in ascending order

C. To reverse the array elements

D. To find the maximum element in the entire array

Solution

1. Step 1: Understand the sliding window concept

   The sliding window technique moves a fixed-size window over the array to calculate values like sums or averages efficiently.

2. Step 2: Identify the purpose

   It avoids recalculating the sum or property for each new window by updating the previous result, making it efficient.

3. Final Answer:

   To efficiently find subarray sums or properties without recalculating everything each time -> Option A

4. Quick Check:

   Sliding window = efficient subarray calculation [OK]

Hint: Think of a moving window updating results fast [OK]

Common Mistakes:

• Confusing sliding window with sorting
• Thinking it reverses arrays
• Assuming it finds max element globally

2. Which of the following is the correct syntax to start a sliding window loop of size k over an array arr in C#?

easy

A. for (int i = 1; i <= arr.Length - k; i++)

B. for (int i = 0; i <= arr.Length - k; i++)

C. for (int i = k; i < arr.Length; i++)

D. for (int i = 0; i < arr.Length + k; i++)

Solution

1. Step 1: Determine valid window start indices

   The window of size k can start at index 0 up to index arr.Length - k to avoid overflow.

2. Step 2: Check loop condition

   The loop must run while i <= arr.Length - k to include the last valid window.

3. Final Answer:

   for (int i = 0; i <= arr.Length - k; i++) -> Option B

4. Quick Check:

   Loop runs from 0 to arr.Length - k [OK]

Hint: Window ends at arr.Length - k index [OK]

Common Mistakes:

• Using < instead of <= causing last window skip
• Starting loop at 1 instead of 0
• Looping beyond array length causing errors

3. What is the output of the following C# code?

int[] arr = {1, 3, 2, 6, -1, 4, 1, 8, 2};
int k = 5;
int maxSum = 0;
int windowSum = 0;

for (int i = 0; i < k; i++)
    windowSum += arr[i];
maxSum = windowSum;

for (int i = k; i < arr.Length; i++) {
    windowSum += arr[i] - arr[i - k];
    if (windowSum > maxSum)
        maxSum = windowSum;
}
Console.WriteLine(maxSum);

medium

A. 16

B. 17

C. 18

D. 20

Solution

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

   Sum of arr[0..4] = 1+3+2+6+(-1) = 11

2. Step 2: Slide window and update sums

   Next windows and sums: - i=5: windowSum = 11 + 4 - 1 = 14 (maxSum=14) - i=6: windowSum = 14 + 1 - 3 = 12 (maxSum=14) - i=7: windowSum = 12 + 8 - 2 = 18 (maxSum=18) - i=8: windowSum = 18 + 2 - 6 = 14 (maxSum=18)

3. Final Answer:

   18 -> Option C

4. Quick Check:

   Max sliding window sum = 18 [OK]

Hint: Update sum by adding new and removing old element [OK]

Common Mistakes:

• Not subtracting the element leaving the window
• Starting loop from wrong index
• Incorrect initial sum calculation

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

int maxSum = 0;
int windowSum = 0;
for (int i = 0; i < arr.Length; i++) {
    windowSum += arr[i];
    if (i >= k) {
        windowSum -= arr[i - k];
        if (windowSum > maxSum)
            maxSum = windowSum;
    }
}
Console.WriteLine(maxSum);

medium

A. maxSum should be initialized to int.MinValue to handle negative sums

B. The if condition should be i > k instead of i >= k

C. The loop should start from i = k instead of 0

D. windowSum should be reset to 0 inside the loop

Solution

1. Step 1: Analyze maxSum initialization

   maxSum is 0 initially, which fails if all sums are negative because maxSum will never update.

2. Step 2: Correct initialization for negative values

   Initialize maxSum to int.MinValue to correctly track maximum sum even if negative.

3. Final Answer:

   maxSum should be initialized to int.MinValue to handle negative sums -> Option A

4. Quick Check:

   maxSum init = int.MinValue for negatives [OK]

Hint: Use int.MinValue for max tracking with negatives [OK]

Common Mistakes:

• Initializing maxSum to 0 causes wrong max for negative arrays
• Changing loop start incorrectly
• Resetting windowSum inside loop breaks sliding window

5. Given an array of integers, write a sliding window approach to find the length of the longest subarray with sum less than or equal to S. Which of the following describes the correct approach?

hard

A. Use a stack to store elements and pop when sum exceeds S

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

C. Calculate sum of all subarrays and pick the longest with sum ≤ S

D. Use two pointers to expand and shrink the window, updating sum and max length accordingly

Solution

1. Step 1: Use two pointers to represent window boundaries

   Start both pointers at the beginning and expand the right pointer to include elements while sum ≤ S.

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

   If sum > S, move left pointer forward subtracting elements until sum ≤ S again, tracking max length.

3. Final Answer:

   Use two pointers to expand and shrink the window, updating sum and max length accordingly -> Option D

4. Quick Check:

   Two-pointer sliding window for variable length subarray [OK]

Hint: Expand right, shrink left to keep sum ≤ S [OK]

Common Mistakes:

• Sorting array breaks original order needed
• Checking all subarrays is inefficient
• Using stack is not suitable for sum constraints