Sliding Window on Arrays in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to slide the window by moving the start index forward.

DSA Go

start := 0
for end := 0; end < len(nums); end++ {
    if end-start+1 > k {
        start = start [1] 1
    }
}

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes the window to move backward.

Using '*' or '/' operators here does not correctly move the window.

To slide the window forward, we increase the start index by 1 using the '+' operator.

2fill in blank

medium

Complete the code to calculate the sum of elements inside the sliding window.

DSA Go

windowSum := 0
start := 0
for end := 0; end < len(nums); end++ {
    windowSum += nums[end]
    if end >= k-1 {
        result = append(result, windowSum)
        windowSum -= nums[[1]]
        start++
    }
}

Drag options to blanks, or click blank then click option'

Ak

Bend

Cstart

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting nums[end] instead of nums[start] causes incorrect sums.

Using a fixed index like 0 or k instead of start leads to errors.

We subtract the element at the start index to slide the window forward correctly.

3fill in blank

hard

Fix the error in the code to correctly find the maximum sum of any sliding window of size k.

DSA Go

maxSum := 0
windowSum := 0
start := 0
for end := 0; end < len(nums); end++ {
    windowSum += nums[end]
    if end-start+1 == k {
        if windowSum > maxSum {
            maxSum = windowSum
        }
        windowSum -= nums[[1]]
        start++
    }
}

Drag options to blanks, or click blank then click option'

Astart

Bend

Ck

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting nums[end] causes wrong window sums.

Using a fixed index like 0 or k causes index errors.

Subtracting nums[start] removes the leftmost element as the window slides.

4fill in blank

hard

Fill both blanks to create a sliding window that finds the minimum element in each window of size k.

DSA Go

result := []int{}
for i := 0; i <= len(nums)-k; i++ {
    minVal := nums[i]
    for j := i; j < i+[1]; j++ {
        if nums[j] [2] minVal {
            minVal = nums[j]
        }
    }
    result = append(result, minVal)
}

Drag options to blanks, or click blank then click option'

Ak

B<

C>

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' finds the maximum, not minimum.

Looping until i+1 or i instead of i+k misses elements.

The inner loop runs from i to i+k to cover the window. We check if nums[j] is less than minVal to find the minimum.

5fill in blank

hard

Fill all three blanks to create a map of word lengths for words longer than 3 characters.

DSA Go

words := []string{"go", "code", "slide", "window"}
lengths := map[string]int{}
for _, word := range words {
    if len(word) [1] 3 {
        lengths[[2]] = [3]
    }
}

Drag options to blanks, or click blank then click option'

A>

Bword

Clen(word)

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' includes shorter words.

Using incorrect keys or values in the map.

We check if the word length is greater than 3, then map the word to its length.

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