Sliding Window on Arrays in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the sliding window technique in arrays?

It is a method to create a 'window' of fixed size that moves over the array to process elements efficiently without re-computing everything from scratch.

beginner

Why is sliding window more efficient than nested loops for fixed-size subarray problems?

Because it reuses previous computations by adding the new element and removing the old one as the window moves, avoiding repeated work.

beginner

In sliding window, what does the 'window size' represent?

The number of consecutive elements considered at a time while moving through the array.

intermediate

How do you update the window when it slides forward by one position?

Remove the element that goes out of the window from the left and add the new element that comes into the window from the right.

beginner

What is a common use case of sliding window on arrays?

Finding the maximum or minimum sum of any subarray of fixed size, or checking for patterns within a moving window.

What does the sliding window technique help to avoid when processing arrays?

A Changing the array size

B Recomputing sums or results for overlapping parts

C Using extra memory for each element

D Sorting the entire array

Sliding window avoids recomputing results for overlapping parts by updating the window incrementally.

If the window size is 3, how many elements are processed at once?

A 1

B 4

C 2

D 3

The window size defines how many consecutive elements are processed at once.

When sliding the window forward by one, what happens to the element leaving the window?

A It is removed from the current calculation

B It is added again

C It stays in the window

D It is doubled

The element leaving the window is removed from the current calculation to update the window result.

Which problem is best solved using sliding window?

A Sorting the array

B Finding the sum of all elements in the array

C Finding the maximum sum of any subarray of fixed size

D Reversing the array

Sliding window efficiently finds maximum sums of fixed-size subarrays by moving the window.

What is the main advantage of sliding window over brute force?

A Avoids repeated calculations

B Runs slower

C Uses more memory

D Requires recursion

Sliding window avoids repeated calculations, making it faster than brute force.

Explain how the sliding window technique works on an array with a fixed window size.

Think about how you slide a small frame over a long picture to see parts at a time.

Describe a real-life example where sliding window can be used to solve a problem.

Imagine checking the temperature for every 3-day period in a week.

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