Divide and Conquer Strategy and Recurrence Relations in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

int mid = (low [1] high) / 2;

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using multiplication instead of addition.

Dividing low by 2 before adding high.

The midpoint is calculated by adding low and high, then dividing by 2.

2fill in blank

medium

Complete the code to correctly express the recurrence relation for merge sort.

DSA C++

T(n) = 2T([1]) + cn

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

An*2

Bn/2

Cn-1

Dn+1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n*2 instead of n/2.

Using n-1 or n+1 which are incorrect splits.

Merge sort divides the problem into two halves of size n/2 each.

3fill in blank

hard

Fix the error in the recurrence relation for binary search.

DSA C++

T(n) = T([1]) + c

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

An/2

Bn*2

Cn-1

Dn+1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n*2 which increases problem size.

Using n-1 or n+1 which are linear reductions.

Binary search reduces the problem size by half each time, so the recurrence is T(n) = T(n/2) + c.

4fill in blank

hard

Fill both blanks to complete the recurrence relation and its base case for quicksort.

DSA C++

T(n) = T([1]) + T([2]) + cn

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

Ak

Bn-k-1

Cn/2

Dn-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using equal halves like n/2 for quicksort which is not always true.

Using n-1 for both parts which is incorrect.

Quicksort divides the array into two parts: one of size k and the other of size n-k-1.

5fill in blank

hard

Fill all three blanks to complete the recurrence relation and the condition for the base case in a divide and conquer algorithm.

DSA C++

if (n [1] 1) return c;\nT(n) = aT(n [2] b) + f(n) where [3] is the division factor

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

A<=

B/

Cb

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= instead of <= for base case.

Using multiplication instead of division for problem size reduction.

Confusing the division factor with the number of subproblems.

The base case is when n is less than or equal to 1. The recurrence divides n by b, where b is the division factor.

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 Divide and Conquer concept

   Divide and conquer splits a problem into smaller parts, solves each part, then merges results.

2. Step 2: Compare options

   Break the problem into smaller subproblems, solve them independently, then combine results correctly describes this process. Options B, C, and D describe other strategies like brute force or dynamic programming.

3. Final Answer:

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

4. Quick Check:

   Divide and Conquer = Break, 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 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) = T(n-1) + O(1)

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

D. T(n) = T(n/2) + O(log n)

Solution

1. Step 1: Recall Merge Sort recurrence

   Merge Sort divides array into two halves, sorts each half (2T(n/2)) and merges in O(n).

2. Step 2: Match with options

   T(n) = 2T(n/2) + O(n) matches 2T(n/2) + O(n). Others describe different algorithms or complexities.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Using T(n-1) instead of T(n/2)
• Confusing merge cost with recursive calls
• Choosing quadratic terms incorrectly

3. What is the output of the following C++ function call:

int count = 0;
int func(int n) {
  if (n <= 1) return 1;
  count++;
  return func(n/2) + func(n/2);
}
int main() {
  func(8);
  std::cout << count;
  return 0;
}

medium

A. 15

B. 7

C. 8

D. 6

Solution

1. Step 1: Understand recursion calls and count increments

   func(8) calls func(4) twice, each func(4) calls func(2) twice, and so on until n <= 1.

2. Step 2: Count increments at each call

   Count increments at each non-base call. The recursion tree has 15 non-base calls (all calls except those where n <= 1). Thus, count increments 15 times.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Count increments for each non-base call = 15 [OK]

Hint: Count increments at each non-base recursive call [OK]

Common Mistakes:

• Counting base case calls
• Not doubling calls for each recursion
• Miscounting due to integer division

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

T(n) = 2T(n/2) + n^2 for n > 1
T(1) = 1

What is the correct time complexity?

medium

A. O(n log n)

B. O(n^2)

C. O(n^2 log n)

D. O(n)

Solution

1. Step 1: Analyze the recurrence relation

   T(n) = 2T(n/2) + n^2 means dividing problem into two halves and combining in n^2 time.

2. Step 2: Apply Master Theorem

   Here, a=2, b=2, log_b a = 1, f(n)=n^2 = Ω(n^{1 + ε}) for some ε > 0 (case 3). Thus, time complexity is Θ(n^2) = O(n^2).

3. Final Answer:

   O(n^2) -> Option B

4. Quick Check:

   Master Theorem with f(n)=n^2 gives O(n^2) [OK]

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

Common Mistakes:

• Assuming O(n log n) without checking f(n)
• Ignoring the power of n in f(n)
• Misapplying Master Theorem cases

5. Given the recurrence relation

T(n) = 3T(n/3) + n / log n

Using the Master Theorem, what is the asymptotic time complexity of T(n)?

hard

A. O(n)

B. O(n^{log_3 3})

C. O(n log n)

D. O(n log log n)

Solution

1. Step 1: Identify parameters for Master Theorem

   a=3, b=3, so n^{log_b a} = n^{log_3 3} = n^1 = n.

2. Step 2: Compare f(n) = n / log n with n^{log_b a} = n

   f(n) grows slower than n by a factor of log n. This is a case where f(n) = n / log n = n^{1} / log n.

3. Step 3: Apply Master Theorem case 2 variant

   Since f(n) = n / log n = n^{1} / log n, which is smaller than n^{1} by a log factor, the complexity is O(n log log n).

4. Final Answer:

   O(n log log n) -> Option D

5. Quick Check:

   Master Theorem with f(n) = n / log n gives O(n log log n) [OK]

Hint: Compare f(n) to n^{log_b a} carefully for log factors [OK]

Common Mistakes:

• Ignoring the log factor in f(n)
• Choosing O(n) or O(n log n) incorrectly
• Miscomputing log base