Sliding Window on Arrays in DSA Go - Execution Trace

Concept Flow - Sliding Window on Arrays

Start: Initialize window boundaries

↓

Expand window to right

↓

Check window size

↓

Expand right

↓

Repeat

↓

Repeat until end

The sliding window moves over the array by expanding right until the window size matches the target, then processes the window and slides from the left to continue.

Execution Sample

DSA Go

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

Prints all subarrays of size 3 by sliding a window over the array.

Execution Table

Step	Operation	Window Start Index	Window End Index	Window Elements	Action
1	Initialize window	0	2	[1, 3, 2]	Print window
2	Slide window right	1	3	[3, 2, 5]	Print window
3	Slide window right	2	4	[2, 5, 1]	Print window
4	Slide window right	3	5	[5, 1, 1]	Print window
5	Slide window right	4	6	[1, 1, 2]	Print window
6	End loop	-	-	-	Reached end of array for window size 3

💡 Window start index reached len(arr) - windowSize, no more full windows possible

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
i (window start)	0	0	1	2	3	4	5 (loop ends)
window	[]	[1,3,2]	[3,2,5]	[2,5,1]	[5,1,1]	[1,1,2]	[]

Key Moments - 3 Insights

Why does the loop stop when i reaches len(arr) - windowSize?

Because at that point, the window from i to i+windowSize fits exactly inside the array. Beyond that, the window would go out of bounds. See execution_table row 6.

Why do we slice arr[i : i+windowSize] instead of arr[i : i+windowSize+1]?

Because slicing in Go excludes the end index. So arr[i : i+windowSize] includes exactly windowSize elements. Refer to execution_table rows 1-5 for window elements.

What happens if windowSize is larger than the array length?

The loop condition i <= len(arr)-windowSize becomes false immediately, so no windows are processed. This prevents out-of-range errors.

Visual Quiz - 3 Questions

Look at the execution_table, what is the window elements at Step 3?

A [2, 5, 1]

B [3, 2, 5]

C [5, 1, 1]

D [1, 3, 2]

💡 Hint

Check the 'Window Elements' column at Step 3 in execution_table.

At which step does the window start index become 4?

A Step 2

B Step 3

C Step 5

D Step 4

💡 Hint

Look at the 'Window Start Index' column in execution_table rows.

If windowSize was 4, how many times would the loop run?

A 3 times

B 4 times

C 5 times

D 6 times

💡 Hint

Calculate len(arr) - windowSize + 1 with arr length 7 and windowSize 4.

Concept Snapshot

Sliding Window on Arrays:
- Move a fixed-size window over array indices
- Window covers arr[i : i+windowSize]
- Loop i from 0 to len(arr)-windowSize
- Process each window (e.g., sum, max, print)
- Efficient for contiguous subarray problems

Full Transcript

Sliding window on arrays means moving a fixed-size segment over the array step by step. We start with the window at the beginning, then slide it right one element at a time until the window reaches the end. At each step, we look at the elements inside the window. The loop stops when the window would go past the array end. This method helps process all subarrays of a certain size efficiently.

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