Sliding Window on Arrays in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the window sum with the first k elements.

DSA Rust

let window_sum: i32 = nums.iter().take([1]).copied().sum();

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

Ak

Bnums.len()

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums.len() instead of k causes summing the whole array.

Using 0 or 1 as the count leads to incorrect window size.

We take the first k elements to initialize the window sum for the sliding window.

2fill in blank

medium

Complete the code to slide the window by removing the left element and adding the right element.

DSA Rust

window_sum = window_sum - nums[i - [1]] + nums[i];

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

Ai

B1

Ck

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting nums[i - 1] instead of nums[i - k] shifts the window incorrectly.

Using i or 0 instead of k causes wrong index calculation.

We subtract the element leaving the window at index i-k and add the new element at index i.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over the sliding windows.

DSA Rust

for i in [1]..nums.len() {

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

Ak

B0

C1

Dnums.len() - k

Attempts:

3 left

💡 Hint

Common Mistakes

Starting at 0 causes index underflow when subtracting k.

Starting at 1 or nums.len() - k misses windows or causes errors.

The loop starts from k to slide the window from the end of the first window to the last element.

4fill in blank

hard

Fill both blanks to create a vector of maximum sums for each sliding window.

DSA Rust

let mut max_sums = vec![nums.iter().take([1]).copied().sum()];
for i in [2]..nums.len() {
    // sliding window update
}

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

Ak

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 instead of k causes incorrect initialization or loop start.

Mixing different values for the two blanks breaks the logic.

Initialize max_sums with the sum of the first k elements and start the loop from k to slide the window.

5fill in blank

hard

Fill both blanks to update the window sum, track the max sum, and push it to the result vector.

DSA Rust

window_sum = window_sum - nums[i - [1]] + nums[i];
max_sum = max_sum.max(window_sum);
max_sums.push([2]);

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

Ak

Bmax_sum

Cwindow_sum

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing window_sum instead of max_sum causes incorrect results.

Using wrong index subtraction causes runtime errors.

Subtract the element leaving the window at i-k, update max_sum, and push max_sum to max_sums 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