Sliding Window Maximum Using Deque in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the deque index for the first element.

DSA C

deque[0] = [1];

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

A0

B1

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 for the first index.

Initializing deque with -1 which is invalid index.

The deque should start with the index 0 of the first element.

2fill in blank

medium

Complete the condition to remove indices from the back of the deque if current element is greater.

DSA C

while (rear >= front && arr[i] [1] arr[deque[rear]]) {
    rear--;
}

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

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' causing wrong removals.

Using '>=' which may remove equal elements unnecessarily.

We remove from the back while current element is greater than elements at those indices.

3fill in blank

hard

Fix the error in the condition to remove indices out of the current window from the front.

DSA C

if (deque[front] [1] i - k + 1) {
    front++;
}

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' removes valid indices too early.

Using '>=' or '>' reverses the logic.

We remove indices strictly less than the window start index.

4fill in blank

hard

Fill both blanks to correctly assign the maximum for the current window and update the deque.

DSA C

max_values[i - k + 1] = arr[deque[[1]]];
deque[++rear] = [2];

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

Afront

Brear

Ci

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using rear instead of front for max index.

Adding wrong index to deque.

Maximum is at deque front; add current index i to deque rear.

5fill in blank

hard

Fill all three blanks to complete the sliding window maximum function using deque.

DSA C

for (int i = 1; i < n; i++) {
    while (rear >= front && arr[i] [1] arr[deque[rear]]) {
        rear--;
    }
    deque[++rear] = [2];
    if (deque[front] [3] i - k + 1) {
        front++;
    }
    if (i >= k - 1) {
        max_values[i - k + 1] = arr[deque[front]];
    }
}

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

A>

Bi

C<

Dfront

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators causing incorrect deque updates.

Adding wrong index to deque.

Incorrect condition to remove out-of-window indices.

Remove smaller elements with '>'; add current index i; remove indices less than window start with '<'.

Practice

1. What is the main purpose of using a deque in the Sliding Window Maximum problem?

easy

A. To store the sum of elements in each window

B. To sort the array elements before processing

C. To reverse the array for easier traversal

D. To store indices of elements in a way that helps find the maximum in O(1) time per window

Solution

1. Step 1: Understand the role of deque

   A deque stores indices of elements in a way that the front always has the index of the maximum element for the current window.

2. Step 2: Why deque helps in O(1) max retrieval

   Because the deque maintains elements in decreasing order, the maximum is always at the front, allowing quick access without scanning the window.

3. Final Answer:

   To store indices of elements in a way that helps find the maximum in O(1) time per window -> Option D

4. Quick Check:

   Deque stores max indices for O(1) max retrieval [OK]

Hint: Deque keeps max at front for quick max retrieval [OK]

Common Mistakes:

• Thinking deque stores actual values instead of indices
• Using deque to sort elements
• Using deque to store sums

2. Which of the following is the correct way to declare a deque in C for storing integer indices?

easy

A. struct deque dq; // assuming a deque struct is defined

B. deque<int> dq;

C. int dq[];

D. deque dq = new deque();

Solution

1. Step 1: Understand C data structure declaration

   C does not have built-in deque type; we use structs or arrays to implement it.

2. Step 2: Identify correct syntax

   struct deque dq; // assuming a deque struct is defined shows a struct declaration which is the correct way in C if a deque struct is defined.

3. Final Answer:

   struct deque dq; // assuming a deque struct is defined -> Option A

4. Quick Check:

   C uses structs for custom data types [OK]

Hint: Use struct for deque in C, no built-in deque type [OK]

Common Mistakes:

• Using C++ syntax like deque<int>
• Trying to declare deque like an array without size
• Using Java or other language syntax

3. What is the output of the sliding window maximum for array [2, 1, 3, 4, 6, 3, 8, 9, 10, 12, 56] with window size 4?

medium

A. [3, 4, 6, 6, 8, 9, 10, 12, 56]

B. [2, 3, 4, 6, 8, 9, 10, 12]

C. [4, 6, 6, 8, 9, 10, 12, 56]

D. [6, 6, 8, 9, 10, 12, 56]

Solution

1. Step 1: Slide window and find max for each window of size 4

   Windows and max values:
   Window 1: [2,1,3,4] max=4
   Window 2: [1,3,4,6] max=6
   Window 3: [3,4,6,3] max=6
   Window 4: [4,6,3,8] max=8
   Window 5: [6,3,8,9] max=9
   Window 6: [3,8,9,10] max=10
   Window 7: [8,9,10,12] max=12
   Window 8: [9,10,12,56] max=56

2. Step 2: Collect all max values in order

   The max values are [4, 6, 6, 8, 9, 10, 12, 56]

3. Final Answer:

   [4, 6, 6, 8, 9, 10, 12, 56] -> Option C

4. Quick Check:

   Max per window = [4,6,6,8,9,10,12,56] [OK]

Hint: Slide window, track max with deque front [OK]

Common Mistakes:

• Including partial windows smaller than size 4
• Off-by-one errors in window indexing
• Confusing max values with sums

4. Given this snippet for sliding window maximum, what is the bug?

for (int i = 0; i < n; i++) {
  while (!dq.empty() && arr[dq.back()] <= arr[i])
    dq.pop_back();
  dq.push_back(i);
  if (dq.front() <= i - k)
    dq.pop_front();
  if (i >= k - 1)
    printf("%d ", arr[dq.front()]);
}

medium

A. The condition to pop front should be < instead of <=

B. No bug, code is correct

C. The condition to pop front should be checked before pushing current index

D. The order of popping from back and pushing is wrong

Solution

1. Step 1: Analyze deque maintenance logic

   The code pops smaller elements from back before pushing current index, ensuring decreasing order.

2. Step 2: Check window boundary condition

   The code pops front if the front index is out of current window (<= i-k), which is correct.

3. Step 3: Confirm output printing

   Printing starts after first full window (i >= k-1), printing max at front index.

4. Final Answer:

   No bug, code is correct -> Option B

5. Quick Check:

   Deque logic and conditions are correct [OK]

Hint: Check deque front removal after pushing current index [OK]

Common Mistakes:

• Removing front before pushing current index
• Using wrong comparison operators in conditions
• Printing output before full window

5. How would you modify the sliding window maximum algorithm using deque to find the sliding window minimum instead?

hard

A. Change the comparison from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] when popping from back

B. Reverse the array before applying the same algorithm

C. Use a stack instead of a deque

D. Change the window size to half

Solution

1. Step 1: Understand max vs min logic in deque

   For max, we pop from back while elements are smaller or equal. For min, we pop while elements are greater or equal.

2. Step 2: Modify comparison operator

   Change condition from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] to maintain increasing order for min.

3. Final Answer:

   Change the comparison from arr[dq.back()] <= arr[i] to arr[dq.back()] >= arr[i] when popping from back -> Option A

4. Quick Check:

   Reverse comparison for min sliding window [OK]

Hint: Flip comparison sign to find min instead of max [OK]

Common Mistakes:

• Reversing array instead of changing comparison
• Using stack which doesn't maintain order
• Changing window size unnecessarily