Divide and Conquer Strategy and Recurrence Relations in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the midpoint for dividing the array.

DSA Javascript

const mid = Math.floor((start + [1]) / 2);

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

Aend

Blength

C0

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using length instead of end index.

Using start twice in the calculation.

The midpoint is calculated by averaging the start and end indices.

2fill in blank

medium

Complete the code to recursively call the function on the left half of the array.

DSA Javascript

return divideAndConquer(arr, start, [1]);

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

Amid + 1

Bend

Cstart + 1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 which skips the midpoint.

Using end which covers the whole array.

The left half ends at the midpoint index.

3fill in blank

hard

Fix the error in the base case condition to stop recursion correctly.

DSA Javascript

if (start [1] end) return arr[start];

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

A===

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than or less than which causes wrong stopping.

Using greater or equal which may cause infinite recursion.

The base case is when start equals end, meaning one element left.

4fill in blank

hard

Fill both blanks to correctly combine the results from left and right halves.

DSA Javascript

const left = divideAndConquer(arr, start, [1]);
const right = divideAndConquer(arr, [2], end);

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

Amid

Bmid + 1

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using start for right half start index.

Using end for left half end index.

The left half ends at mid, the right half starts at mid + 1.

5fill in blank

hard

Fill all three blanks to write the recurrence relation for the time complexity.

DSA Javascript

T(n) = 2T(n [1] 2) [2] c [3] n

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

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for problem size.

Using subtraction instead of addition for combining costs.

The recurrence relation for divide and conquer is T(n) = 2T(n/2) + cn.

Practice

1. What is the main idea behind the Divide and Conquer strategy in algorithms?

easy

A. Solve the problem by trying all possible solutions one by one

B. Use a single loop to iterate through all elements

C. Break the problem into smaller subproblems, solve them independently, then combine results

D. Store all intermediate results to avoid repeated calculations

Solution

1. Step 1: Understand the divide and conquer approach

   It splits a big problem into smaller parts that are easier to solve.

2. Step 2: Combine the solutions of smaller parts

   After solving subproblems, their results are merged to get the final answer.

3. Final Answer:

   Break the problem into smaller subproblems, solve them independently, then combine results -> Option C

4. Quick Check:

   Divide and conquer = split, solve, combine [OK]

Hint: Divide and conquer means split, solve, then merge results [OK]

Common Mistakes:

• Confusing divide and conquer with brute force
• Thinking it uses only one loop
• Mixing it with dynamic programming

2. Which of the following is the correct syntax for the recurrence relation of Merge Sort?

easy

A. T(n) = 3T(n/3) + O(log n)

B. T(n) = 2T(n/2) + O(n)

C. T(n) = T(n/2) + O(n^2)

D. T(n) = T(n-1) + O(1)

Solution

1. Step 1: Recall Merge Sort recurrence

   Merge Sort splits array into two halves, sorts each half, then merges in O(n).

2. Step 2: Write recurrence form

   It calls itself twice on half size (2T(n/2)) plus linear merge time O(n).

3. Final Answer:

   T(n) = 2T(n/2) + O(n) -> Option B

4. Quick Check:

   Merge Sort recurrence = 2T(n/2) + n [OK]

Hint: Merge Sort splits in two, merges in linear time [OK]

Common Mistakes:

• Using T(n-1) instead of T(n/2)
• Adding O(n^2) instead of O(n)
• Confusing with Quick Sort recurrence

3. What is the output of the following JavaScript function that uses divide and conquer?

function sumArray(arr) {
  if (arr.length === 1) return arr[0];
  const mid = Math.floor(arr.length / 2);
  return sumArray(arr.slice(0, mid)) + sumArray(arr.slice(mid));
}
console.log(sumArray([2, 4, 6, 8]));

medium

A. 20

B. 24

C. 10

D. Error

Solution

1. Step 1: Understand the function behavior

   The function splits the array into halves recursively and sums elements.

2. Step 2: Trace the calls for [2,4,6,8]

   Split into [2,4] and [6,8], sumArray([2,4]) = 6, sumArray([6,8]) = 14, total 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Sum of 2+4+6+8 = 20 [OK]

Hint: Divide array, sum halves, add results [OK]

Common Mistakes:

• Confusing slice indices causing wrong sums
• Assuming function returns array instead of sum
• Forgetting base case for single element

4. Identify the error in this divide and conquer function to find the maximum element:

function findMax(arr) {
  if (arr.length === 1) return arr[0];
  const mid = Math.floor(arr.length / 2);
  const leftMax = findMax(arr.slice(0, mid));
  const rightMax = findMax(arr.slice(mid));
  return leftMax > rightMax ? leftMax : rightMax;
}
console.log(findMax([]));

medium

A. Fails on empty array input causing infinite recursion

B. Incorrect comparison operator used

C. No error, works correctly

D. Slices do not split array properly

Solution

1. Step 1: Check base case for empty array

   The function does not handle empty array input, so arr.length === 0 is not checked.

2. Step 2: Understand recursion on empty array

   Calling findMax([]) leads to infinite recursion because base case is only for length 1.

3. Final Answer:

   Fails on empty array input causing infinite recursion -> Option A

4. Quick Check:

   Empty input needs base case to avoid infinite calls [OK]

Hint: Always check for empty input base case [OK]

Common Mistakes:

• Ignoring empty array edge case
• Assuming slice always returns non-empty array
• Thinking comparison operator is wrong

5. Given the recurrence relation T(n) = 3T(n/3) + n, what is the time complexity using the Master Theorem?

hard

A. O(log n)

B. O(n^2)

C. O(n)

D. O(n log n)

Solution

1. Step 1: Identify parameters for Master Theorem

   Here, a=3, b=3, f(n)=n. Calculate log_b(a) = log_3(3) = 1.

2. Step 2: Compare f(n) with n^log_b(a)

   f(n) = n and n^log_b(a) = n^1 = n, so f(n) = Theta(n^log_b(a)).

3. Step 3: Apply Master Theorem case 2

   When f(n) = Theta(n^log_b(a)), T(n) = Theta(n log n).

4. Final Answer:

   O(n log n) -> Option D

5. Quick Check:

   Master Theorem case 2 = n log n [OK]

Hint: Compare f(n) to n^log_b(a) for Master Theorem [OK]

Common Mistakes:

• Confusing log base or parameters a and b
• Choosing wrong Master Theorem case
• Ignoring the extra n term in recurrence