Sliding Window on Arrays in DSA Go - 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.

func maxSumSubarray(arr []int, k int) int {
    maxSum := 0
    windowSum := 0
    for i := 0; i < k; i++ {
        windowSum += arr[i]
    }
    maxSum = windowSum
    for i := k; i < len(arr); 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 over the array elements.
• 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: Doubling the input roughly doubles the work done.

Final Time Complexity

Time Complexity: O(n)

This means the time grows linearly with the size of the input array.

Common Mistake

[X] Wrong: "The sliding window has nested loops, so it must be O(n²)."

[OK] Correct: The window moves forward by one step each time without redoing all work, so the total steps add up to n, not n squared.

Interview Connect

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

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 on arrays?

easy

A. To efficiently process subarrays of fixed size without re-computing everything

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 sliding window concept

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

2. Step 2: Identify purpose

   It avoids recalculating values for overlapping parts of subarrays, improving performance.

3. Final Answer:

   To efficiently process subarrays of fixed size without re-computing everything -> Option A

4. Quick Check:

   Sliding window = efficient subarray processing [OK]

Hint: Sliding window avoids repeated work on overlapping parts [OK]

Common Mistakes:

• Confusing sliding window with sorting
• Thinking it reverses arrays
• Assuming it finds max of whole array

2. Which of the following Go code snippets correctly initializes a sliding window sum for the first 3 elements of an integer array arr?

easy

A. sum := 0
for i := 0; i < 3; i++ {
sum += arr[i]
}

B. sum := 0
for i := 1; i <= 3; i++ {
sum += arr[i]
}

C. sum := 0
for i := 0; i <= 3; i++ {
sum += arr[i]
}

D. sum := 0
for i := 1; i < 3; i++ {
sum += arr[i]
}

Solution

1. Step 1: Check loop bounds for first 3 elements

   Array indices start at 0, so first 3 elements are at indices 0,1,2. Loop must run i from 0 to less than 3.

2. Step 2: Verify each option

   sum := 0 for i := 0; i < 3; i++ { sum += arr[i] } loops i=0 to 2 correctly. Others either start at 1 or go out of bounds.

3. Final Answer:

   sum := 0 for i := 0; i < 3; i++ { sum += arr[i] } -> Option A

4. Quick Check:

   Loop i=0 to 2 sums first 3 elements [OK]

Hint: Array indices start at 0; loop to i < window size [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < causing out of bounds
• Incorrect loop end index

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

arr := []int{2, 1, 5, 1, 3, 2}
windowSize := 3
sum := 0
for i := 0; i < windowSize; i++ {
    sum += arr[i]
}
result := []int{sum}
for i := windowSize; i < len(arr); i++ {
    sum += arr[i] - arr[i-windowSize]
    result = append(result, sum)
}
fmt.Println(result)

medium

A. [8 6 9 7]

B. [8 7 9 6 5]

C. [8 7 9 6]

D. [2 1 5 1 3 2]

Solution

1. Step 1: Calculate initial window sum

   First 3 elements: 2+1+5=8. result = [8]

2. Step 2: Slide window and update sums

   i=3: sum=8 + arr[3](1) - arr[0](2) = 8+1-2=7; result=[8,7] i=4: sum=7 + arr[4](3) - arr[1](1) = 7+3-1=9; result=[8,7,9] i=5: sum=9 + arr[5](2) - arr[2](5) = 9+2-5=6; result=[8,7,9,6]

3. Final Answer:

   [8 7 9 6] -> Option C

4. Quick Check:

   Sliding sums = [8 7 9 6] [OK]

Hint: Add new element, subtract old element to update sum [OK]

Common Mistakes:

• Not subtracting the element leaving the window
• Appending sum before updating
• Incorrect initial sum calculation

4. Identify the bug in this Go code that tries to find the maximum sum of any subarray of size 2:

arr := []int{1, 3, 2, 5, 1}
windowSize := 2
maxSum := 0
sum := 0
for i := 0; i < windowSize; i++ {
    sum += arr[i]
}
for i := windowSize; i < len(arr); i++ {
    sum += arr[i] - arr[i-windowSize]
    if sum > maxSum {
        maxSum = sum
    }
}
fmt.Println(maxSum)

medium

A. maxSum should be initialized to a very small number, not 0

B. The first window sum is not compared to maxSum

C. The loop should start from i = 0 in the second loop

D. The subtraction in sum update is incorrect

Solution

1. Step 1: Check initial maxSum and first window sum

   maxSum is 0, sum is sum of first 2 elements (1+3=4). But maxSum is not updated with this initial sum.

2. Step 2: Identify missing comparison

   The first window sum should be compared to maxSum before sliding the window, else maxSum remains 0 if all sums are less.

3. Final Answer:

   The first window sum is not compared to maxSum -> Option B

4. Quick Check:

   Compare initial sum to maxSum before loop [OK]

Hint: Always compare initial window sum to max before sliding [OK]

Common Mistakes:

• Not updating maxSum with initial sum
• Wrong loop start index
• Incorrect sum update formula

5. Given an array of integers, write a Go function using sliding window to find the length of the longest subarray with sum less than or equal to a target value k. Which approach correctly solves this problem?

hard

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

B. Use a stack to store sums and pop when sum exceeds k

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

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

Solution

1. Step 1: Understand problem constraints

   We want the longest subarray with sum ≤ k, which requires dynamic window size adjustment.

2. Step 2: Choose sliding window with two pointers

   Expand right pointer adding elements to sum; if sum > k, move left pointer to shrink window until sum ≤ k; track 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 adjusts size for sum ≤ k [OK]

Hint: Expand right, shrink left when sum exceeds target [OK]

Common Mistakes:

• Sorting array breaks subarray order
• Checking all subarrays is inefficient
• Using stack is not suitable here