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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to correctly divide the problem size by 2 in the recursive call.

DSA C

int divideAndConquer(int n) {
    if (n <= 1) return n;
    return divideAndConquer(n [1] 2) + n;
}

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

A-

B+

C/

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or subtraction instead of division to reduce problem size.

Multiplying n instead of dividing.

The divide and conquer strategy often splits the problem size in half, so dividing n by 2 is correct.

2fill in blank

medium

Complete the recurrence relation formula for the time complexity T(n) = 2 * T(n {{BLANK_1}} 2) + n.

DSA C

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

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

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of division in the recurrence relation.

Using addition which increases problem size.

The recurrence relation divides the problem size by 2, so the operator is division '/'.

3fill in blank

hard

Fix the error in the recursive function call to correctly implement divide and conquer.

DSA C

int solve(int n) {
    if (n <= 1) return 1;
    return 2 * solve([1]) + n;
}

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

An * 2

Bn / 2

Cn + 2

Dn - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Multiplying or adding to n instead of dividing.

Using n - 1 which reduces size by 1, not half.

The problem size should be divided by 2 in the recursive call to correctly apply divide and conquer.

4fill in blank

hard

Fill both blanks to complete the recurrence relation and base case for a divide and conquer algorithm.

DSA C

if (n [1] 1) return 1;
T(n) = 3 * T(n [2] 3) + n;

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

A<=

B>

C/

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than in base case condition.

Using subtraction instead of division in recurrence.

The base case checks if n is less than or equal to 1, and the recurrence divides n by 3.

5fill in blank

hard

Fill all three blanks to complete the recursive function and recurrence relation for a divide and conquer algorithm.

DSA C

int dcFunc(int n) {
    if (n [1] 1) return 1;
    return 4 * dcFunc(n [2] 2) + n * n;
}

T(n) = 4 * T(n [3] 2) + n * n;

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

A<=

B-

C/

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or subtraction instead of division.

Incorrect base case condition.

The base case uses <= 1, and the recursive calls divide n by 2 in both function and recurrence.

Practice

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

easy

A. Use a single loop to solve the problem iteratively

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

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 Divide and Conquer approach

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

2. Step 2: Combine solutions of subproblems

   After solving smaller parts, their results are combined 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 = Break, Solve, Combine [OK]

Hint: Divide big problem into parts, solve parts, then combine [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 on an array of size n?

easy

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

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

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

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

Solution

1. Step 1: Recall Merge Sort process

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

2. Step 2: Write recurrence relation

   It calls itself twice on size n/2 plus linear merge step, so T(n) = 2T(n/2) + O(n).

3. Final Answer:

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

4. Quick Check:

   Merge Sort recurrence = 2 halves + merge O(n) [OK]

Hint: Two halves plus linear merge step means 2T(n/2)+n [OK]

Common Mistakes:

• Using T(n-1) instead of T(n/2)
• Ignoring the merge step cost
• Confusing with quicksort recurrence

3. Consider the following C code implementing a divide and conquer function:

int func(int n) {
    if (n <= 1) return 1;
    return func(n/2) + func(n/2) + n;
}

What is the time complexity T(n) of this function?

medium

A. O(n^2)

B. O(n)

C. O(log n)

D. O(n log n)

Solution

1. Step 1: Identify recurrence relation

   The function calls itself twice on n/2 plus does O(n) work, so T(n) = 2T(n/2) + n.

2. Step 2: Solve recurrence using Master Theorem

   For T(n) = 2T(n/2) + n, the solution is O(n log n).

3. Final Answer:

   O(n log n) -> Option D

4. Quick Check:

   2 calls on half + linear work = n log n [OK]

Hint: Two halves plus linear work means n log n time [OK]

Common Mistakes:

• Thinking it is O(n) by ignoring recursive calls
• Confusing with O(log n) for single call
• Mistaking it for quadratic O(n^2)

4. The following recurrence relation is given:

T(n) = 3T(n/3) + n^2

There is a mistake in the code implementing this recurrence. Which fix will correct the recurrence to match the intended divide and conquer approach?

medium

A. Change the recursive calls to 3T(n/3) and keep n^2 as extra work

B. Change the extra work from n^2 to n log n

C. Change the recursive calls to T(n-1) and keep n^2

D. Change the recursive calls to 2T(n/2) and keep n^2

Solution

1. Step 1: Understand the recurrence meaning

   3T(n/3) means dividing problem into 3 equal parts recursively, plus n^2 extra work.

2. Step 2: Fix code to match recurrence

   Ensure code calls itself 3 times on n/3 size and does n^2 work outside recursion.

3. Final Answer:

   Change the recursive calls to 3T(n/3) and keep n^2 as extra work -> Option A

4. Quick Check:

   3 recursive calls on n/3 + n^2 work = 3T(n/3)+n^2 [OK]

Hint: Match recursive calls count and size to recurrence exactly [OK]

Common Mistakes:

• Changing extra work without fixing recursive calls
• Using wrong subproblem size like n-1
• Reducing calls to 2 instead of 3

5. You have a problem of size n that you divide into 4 subproblems each of size n/2. The extra work done outside the recursive calls is O(n^2). What is the time complexity T(n) of this divide and conquer algorithm?

hard

A. O(n^2 log n)

B. O(n^3)

C. O(n^2)

D. O(n log n)

Solution

1. Step 1: Write the recurrence relation

   Since there are 4 subproblems each size n/2 and extra work n^2, T(n) = 4T(n/2) + n^2.

2. Step 2: Apply Master Theorem

   Here, a=4, b=2, f(n)=n^2. Calculate log_b(a) = log_2(4) = 2. Since f(n) = Theta(n^2), it matches n^{log_b(a)}.

3. Step 3: Determine complexity

   By Master Theorem case 2, T(n) = Theta(n^2 log n).

4. Final Answer:

   O(n^2 log n) -> Option A

5. Quick Check:

   4 calls on half + n^2 work = n^2 log n [OK]

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

Common Mistakes:

• Ignoring the log factor in complexity
• Confusing number of subproblems with size
• Mistaking it for cubic or linear complexity