Sliding Window on Arrays in DSA Go - Build from Scratch

Sliding Window on Arrays

📖 Scenario: You are analyzing daily temperatures recorded over a week. You want to find the highest temperature in every 3-day window to understand short-term heat trends.

🎯 Goal: Build a program that uses the sliding window technique to find the maximum temperature in every 3-day window from a list of daily temperatures.

📋 What You'll Learn

Create a slice called temperatures with exactly these values: 73, 74, 75, 71, 69, 72, 76, 73

Create an integer variable called windowSize and set it to 3

Use a for loop with variable i to slide over temperatures and find the maximum in each window

Store each maximum in a slice called maxTemps

Print the maxTemps slice at the end

💡 Why This Matters

🌍 Real World

Sliding window technique is used in weather data analysis, stock price monitoring, and network traffic analysis to find trends over fixed intervals.

💼 Career

Understanding sliding window helps in roles like data analysis, software development, and system monitoring where efficient data processing is required.

1

Create the temperatures data

Create a slice called temperatures with these exact values: 73, 74, 75, 71, 69, 72, 76, 73

DSA Go

# Create the temperatures slice with the given values
# Your code here

Hint

Use temperatures := []int{73, 74, 75, 71, 69, 72, 76, 73} to create the slice.

2

Set the window size

Create an integer variable called windowSize and set it to 3

DSA Go

package main

func main() {
    temperatures := []int{73, 74, 75, 71, 69, 72, 76, 73}
    // Set the window size to 3
    // Your code here
}

Hint

Use windowSize := 3 to set the size of the sliding window.

3

Find maximum temperatures in each window

Use a for loop with variable i to slide over temperatures from index 0 to len(temperatures) - windowSize. Inside the loop, find the maximum temperature in the current window and append it to a slice called maxTemps. Initialize maxTemps as an empty slice of integers before the loop.

DSA Go

package main

func main() {
    temperatures := []int{73, 74, 75, 71, 69, 72, 76, 73}
    windowSize := 3
    // Initialize maxTemps slice
    // Use a for loop with i to slide over temperatures
    // Find max in each window and append to maxTemps
    // Your code here
}

Hint

Use two loops: outer loop slides the window, inner loop finds max in the window.

4

Print the maximum temperatures slice

Print the maxTemps slice using fmt.Println(maxTemps)

DSA Go

package main

import "fmt"

func main() {
    temperatures := []int{73, 74, 75, 71, 69, 72, 76, 73}
    windowSize := 3
    maxTemps := []int{}

    for i := 0; i  max {
                max = temperatures[j]
            }
        }
        maxTemps = append(maxTemps, max)
    }
    // Print the maxTemps slice
    // Your code here
}

Hint

Use fmt.Println(maxTemps) to print the slice.

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