Sliding Window on Arrays in DSA Rust - Execution Trace

Concept Flow - Sliding Window on Arrays

Initialize window start and end indices

↓

Expand window end

↓

Check window size == target size?

↓

Expand window end

↓

Slide window start forward

↓

Repeat until end of array

We move a window over the array by expanding the end index until the window size matches the target, then process and slide the start forward to continue.

Execution Sample

DSA Rust

let arr = [1, 3, 2, 5, 4];
let k = 3;
for i in 0..=arr.len() - k {
    let window = &arr[i..i+k];
    println!("{:?}", window);
}

Prints all subarrays of size k=3 by sliding a window over the array.

Execution Table

Step	Operation	Window Start	Window End	Window Size	Window Elements	Output
1	Initialize window	0	0	1	[1]	No output, window size < k
2	Expand window end	0	1	2	[1, 3]	No output, window size < k
3	Expand window end	0	2	3	[1, 3, 2]	Print [1, 3, 2]
4	Slide window start	1	3	3	[3, 2, 5]	Print [3, 2, 5]
5	Slide window start	2	4	3	[2, 5, 4]	Print [2, 5, 4]
6	Slide window start	3	5	2	[5, 4]	Stop, window size < k, end reached

💡 Window end reached array length; no more full windows of size k=3

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6
window_start	0	0	0	0	1	2	3
window_end	0	0	1	2	3	4	5
window_size	1	1	2	3	3	3	2
window_elements	[]	[1]	[1, 3]	[1, 3, 2]	[3, 2, 5]	[2, 5, 4]	[5, 4]

Key Moments - 3 Insights

Why does the window size start smaller than k and grow?

At steps 1 and 2 in the execution_table, the window expands by moving the end pointer, so the window size grows until it reaches k before processing.

Why do we slide the window start after processing a full window?

At steps 4 and 5, after printing the current window, we move the start forward to slide the window and keep its size constant at k.

Why does the process stop when the window size is less than k at the end?

At step 6, the window end reaches beyond the array length, so the window can't be full size k anymore, so we stop as shown in the exit_note.

Visual Quiz - 3 Questions

Look at the execution_table, what is the window size at Step 3?

A 1

B 2

C 3

D 4

💡 Hint

Check the 'Window Size' column at Step 3 in the execution_table.

At which step does the window start first move forward?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Window Start' column changes in the execution_table.

If k was 2 instead of 3, how many windows would be printed?

A 3

B 4

C 5

D 6

💡 Hint

Count how many subarrays of size 2 fit in the array length 5.

Concept Snapshot

Sliding Window on Arrays:
- Use two pointers: start and end
- Expand end until window size == k
- Process window (e.g., print or sum)
- Slide start forward to move window
- Repeat until end reaches array length

Full Transcript

Sliding window on arrays uses two pointers to create a moving window of fixed size k. We start with both pointers at the beginning. We expand the end pointer until the window size reaches k. Then we process the window, for example by printing its elements. After processing, we slide the start pointer forward by one to move the window forward, keeping the size constant. We repeat this until the end pointer reaches the end of the array and no more full windows of size k can be formed.

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