Sliding Window on Arrays in DSA Rust - 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 Rust 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 vector called temperatures with exactly these values: 30, 32, 31, 35, 33, 34, 36

Create a variable called window_size and set it to 3

Use a for loop with variable i to slide over temperatures and find the maximum in each 3-day window

Print the resulting vector of maximum temperatures for each window

💡 Why This Matters

🌍 Real World

Sliding window technique is used in weather data analysis, stock price monitoring, and real-time sensor data processing to find trends over fixed-size intervals.

💼 Career

Understanding sliding window algorithms is important for roles in data analysis, software development, and systems engineering where efficient data processing is required.

1

Create the temperature data

Create a vector called temperatures with these exact values: 30, 32, 31, 35, 33, 34, 36

DSA Rust

# Create a vector called temperatures with the values 30, 32, 31, 35, 33, 34, 36
# Your code here

Hint

Use let temperatures = vec![30, 32, 31, 35, 33, 34, 36]; to create the vector.

2

Set the sliding window size

Create a variable called window_size and set it to 3

DSA Rust

fn main() {
    let temperatures = vec![30, 32, 31, 35, 33, 34, 36];
    // Create window_size variable and set it to 3
    // Your code here
}

Hint

Use let window_size = 3; to set the window size.

3

Find maximum temperatures in each sliding window

Use a for loop with variable i to slide over temperatures from index 0 to temperatures.len() - window_size. For each window, find the maximum temperature and store it in a vector called max_temps

DSA Rust

fn main() {
    let temperatures = vec![30, 32, 31, 35, 33, 34, 36];
    let window_size = 3;
    let mut max_temps = Vec::new();
    // Use a for loop with i to slide over temperatures and find max in each window
    // Your code here
}

Hint

Use slicing &temperatures[i..i + window_size] and iter().max() to find the max in each window.

4

Print the maximum temperatures vector

Print the vector max_temps to display the maximum temperatures for each 3-day window

DSA Rust

fn main() {
    let temperatures = vec![30, 32, 31, 35, 33, 34, 36];
    let window_size = 3;
    let mut max_temps = Vec::new();
    for i in 0..=temperatures.len() - window_size {
        let window = &temperatures[i..i + window_size];
        let max_in_window = *window.iter().max().unwrap();
        max_temps.push(max_in_window);
    }
    // Print max_temps vector
    // Your code here
}

Hint

Use println!("{:?}", max_temps); to print the vector.

Practice

1. What is the main purpose of the sliding window technique on arrays?

easy

A. To sort the array in ascending order

B. To efficiently process subarrays of fixed size without recomputing everything

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 process subarrays efficiently.

2. Step 2: Identify the purpose

   It avoids recomputing sums or other values for each subarray by updating results incrementally.

3. Final Answer:

   To efficiently process subarrays of fixed size without recomputing everything -> Option B

4. Quick Check:

   Sliding window = efficient fixed-size subarray processing [OK]

Hint: Think of a moving window that updates results quickly [OK]

Common Mistakes:

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

2. Which Rust code snippet correctly initializes a sliding window of size 3 over a vector nums?

easy

A. for i in 0..nums.len() - 2 { let window = &nums[i..i+3]; }

B. for i in 0..nums.len() { let window = &nums[i..i+3]; }

C. for i in 0..nums.len() - 1 { let window = &nums[i..i+3]; }

D. for i in 0..nums.len() - 3 { let window = &nums[i..i+3]; }

Solution

1. Step 1: Determine valid loop range for window size 3

   Window covers indices i, i+1, i+2, so i+3 must be ≤ nums.len(). Thus, i < nums.len() - 2.

2. Step 2: Check each option's range

   for i in 0..nums.len() - 2 { let window = &nums[i..i+3]; } uses 0..nums.len()-2, which is correct. Others are off by one or more.

3. Final Answer:

   for i in 0..nums.len() - 2 { let window = &nums[i..i+3]; } -> Option A

4. Quick Check:

   Loop till len-2 for size 3 window [OK]

Hint: Window end index = start + size; loop till len - size + 1 [OK]

Common Mistakes:

• Using len()-3 which is too small
• Looping till len() causing out-of-bounds
• Incorrect slice ranges causing panic

3. What is the output of this Rust code?

let nums = vec![1, 3, 5, 7, 9];
let k = 3;
let mut sums = Vec::new();
for i in 0..nums.len() - k + 1 {
    let window_sum: i32 = nums[i..i+k].iter().sum();
    sums.push(window_sum);
}
println!("{:?}", sums);

medium

A. [3, 5, 7]

B. [1, 3, 5]

C. [9, 15, 21]

D. [7, 9, 11]

Solution

1. Step 1: Identify windows and sums

   Windows of size 3 are: [1,3,5], [3,5,7], [5,7,9]. Their sums are 9, 15, 21 respectively.

2. Step 2: Confirm output vector

   The sums vector collects these sums and prints them as [9, 15, 21].

3. Final Answer:

   [9, 15, 21] -> Option C

4. Quick Check:

   Sum windows = [9, 15, 21] [OK]

Hint: Sum each 3-element slice, slide by one [OK]

Common Mistakes:

• Summing wrong slices
• Off-by-one in loop range
• Confusing sums with elements

4. Find the error in this Rust code that tries to print max sum of sliding window size 2:

let nums = vec![4, 2, 1, 7];
let k = 2;
let mut max_sum = 0;
for i in 0..nums.len() - k {
    let sum: i32 = nums[i..i+k].iter().sum();
    if sum > max_sum {
        max_sum = sum;
    }
}
println!("{}", max_sum);

medium

A. Window size k should be 3 instead of 2

B. max_sum should be initialized to i32::MIN instead of 0

C. Sum calculation should use fold instead of iter().sum()

D. Loop range should be 0..nums.len() - k + 1 to include last window

Solution

1. Step 1: Check loop range for window size 2

   Windows are nums[i..i+2], so i must go up to nums.len() - 2 inclusive, meaning 0..nums.len()-1 or 0..nums.len()-k+1.

2. Step 2: Identify off-by-one error

   Current loop ends at nums.len()-k exclusive, missing last window. Correct range is 0..nums.len()-k+1.

3. Final Answer:

   Loop range should be 0..nums.len() - k + 1 to include last window -> Option D

4. Quick Check:

   Loop till len-k+1 for full windows [OK]

Hint: Loop till len - window size + 1 to cover all windows [OK]

Common Mistakes:

• Using wrong loop range causing missed windows
• Initializing max_sum to 0 causing wrong max if negatives
• Unnecessary changes to sum calculation

5. Given an array nums and window size k, how can you find the maximum sum of any subarray of size k using sliding window in Rust efficiently?

hard

A. Calculate sum of first k elements, then slide window by adding next element and subtracting first element of previous window, updating max sum each time

B. Sort the array and sum the last k elements to get maximum sum

C. Use nested loops to sum every subarray of size k and track maximum sum

D. Reverse the array and sum first k elements for maximum sum

Solution

1. Step 1: Compute initial window sum

   Sum the first k elements to get the starting window sum.

2. Step 2: Slide window and update sums

   For each next position, add the new element and subtract the element leaving the window, updating max sum if needed.

3. Step 3: Avoid recomputing sums fully

   This approach avoids nested loops and is efficient with O(n) time complexity.

4. Final Answer:

   Calculate sum of first k elements, then slide window by adding next element and subtracting first element of previous window, updating max sum each time -> Option A

5. Quick Check:

   Sliding window sum update = efficient max sum [OK]

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

Common Mistakes:

• Using nested loops causing slow O(n²) time
• Sorting array which changes element order
• Reversing array which doesn't help sum