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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

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

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

Amiddle

Bstart

Clength

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'start' instead of 'end' causes incorrect midpoint.

Using 'length' is invalid here.

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.

DSA Typescript

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

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

Amid

Bmid + 1

Cend

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'mid + 1' skips the midpoint element.

Using 'end' includes 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 Typescript

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 '>' or '<' causes infinite recursion.

Using '!==' is incorrect for base case.

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

4fill in blank

hard

Fill both blanks to complete the recurrence relation function.

DSA Typescript

function T(n) {
  if (n [1] 1) return 1;
  return 2 * T(n [2] 2) + n;
}

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

A<=

B>

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' in base case causes wrong stopping.

Using '-' instead of '/' changes the recurrence.

The base case is when n is less than or equal to 1. The recurrence calls T with n divided by 2.

5fill in blank

hard

Fill all three blanks to complete the merge step in merge sort.

DSA Typescript

while (i < left.length && j < right.length) {
  if (left[i] [1] right[j]) {
    result.push(left[[2]]);
    i[3];
  } else {
    result.push(right[j]);
    j++;
  }
}

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

A<=

B<

Ci

D++

E--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes duplicates to be handled incorrectly.

Using '--' decrements index wrongly.

We compare left[i] & right[j] using '<'. We push left[i] and increment i with '++'.

Practice

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

easy

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

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

C. Use a single loop to iterate through all elements

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 A

4. Quick Check:

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

Hint: Divide big problems into smaller parts, solve, then merge [OK]

Common Mistakes:

• Confusing divide and conquer with brute force
• Thinking it uses only loops without recursion
• Mixing it up with dynamic programming

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

easy

A. T(n) = 3T(n/3) + O(n^2)

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

C. T(n) = T(n/2) + O(1)

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

Solution

1. Step 1: Recall Merge Sort process

   Merge Sort splits array into two halves, sorts each half, then merges.

2. Step 2: Write recurrence relation

   It calls itself twice on half size (2T(n/2)) plus merging cost 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)
• Confusing merge cost with recursive calls
• Choosing cubic or quadratic terms incorrectly

3. What is the output of the following TypeScript function that uses divide and conquer to find the maximum element?

function findMax(arr: number[], left: number, right: number): number {
  if (left === right) return arr[left];
  const mid = Math.floor((left + right) / 2);
  const maxLeft = findMax(arr, left, mid);
  const maxRight = findMax(arr, mid + 1, right);
  return maxLeft > maxRight ? maxLeft : maxRight;
}

const array = [3, 5, 1, 8, 2];
console.log(findMax(array, 0, array.length - 1));

medium

A. 3

B. 5

C. 8

D. 2

Solution

1. Step 1: Understand the function logic

   The function splits the array into halves recursively and compares max values.

2. Step 2: Trace the recursive calls on array [3,5,1,8,2]

   Max in left half [3,5,1] is 5; max in right half [8,2] is 8; final max is 8.

3. Final Answer:

   8 -> Option C

4. Quick Check:

   Max element found by divide and conquer = 8 [OK]

Hint: Divide array, find max in halves, pick bigger [OK]

Common Mistakes:

• Returning min instead of max
• Incorrect mid calculation causing infinite recursion
• Mixing indices causing wrong subarray

4. Identify the error in the following recurrence relation and fix it:

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

This recurrence is supposed to represent the time complexity of a divide and conquer algorithm similar to Merge Sort.

medium

A. Change O(n^2) to O(n) because merging takes linear time

B. Change 2T(n/2) to T(n-1) because it is not dividing

C. Change O(n^2) to O(log n) because merging is logarithmic

D. No error, the recurrence is correct

Solution

1. Step 1: Analyze the merging cost in divide and conquer

   In Merge Sort-like algorithms, merging two halves takes O(n) time, not O(n^2).

2. Step 2: Correct the recurrence relation

   Replace O(n^2) with O(n) to reflect the linear merging cost.

3. Final Answer:

   Change O(n^2) to O(n) because merging takes linear time -> Option A

4. Quick Check:

   Merging cost = O(n), not O(n^2) [OK]

Hint: Merging in divide and conquer is linear, not quadratic [OK]

Common Mistakes:

• Assuming merging is quadratic
• Changing recursive calls incorrectly
• Ignoring the cost of merging step

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

hard

A. O(log n)

B. O(n^2)

C. O(n^{\log_3 3}) = 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_b(a)} log n) = Theta(n log n).

4. Final Answer:

   O(n log n) -> Option D

5. Quick Check:

   Master Theorem case 2 gives O(n log n) [OK]

Hint: Use Master Theorem: compare f(n) with n^{log_b(a)} [OK]

Common Mistakes:

• Ignoring the log factor in case 2
• Confusing log base or parameters a and b
• Choosing linear time incorrectly