Divide and Conquer Strategy and Recurrence Relations in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Merge Sort Call Count

Consider the following Java method that counts the number of recursive calls made during a merge sort on an array of size n. What is the output when calling countMergeSortCalls(8)?

DSA Java

public class MergeSortCounter {
    static int count = 0;
    public static int countMergeSortCalls(int n) {
        if (n <= 1) {
            count++;
            return count;
        }
        count++;
        countMergeSortCalls(n / 2);
        countMergeSortCalls(n / 2);
        return count;
    }

    public static void main(String[] args) {
        System.out.println(countMergeSortCalls(8));
    }
}

A 8

B 15

C 7

D 16

💡 Hint

Think about how many times the array is split until size 1 and how many calls are made at each level.

The merge sort splits the array into halves recursively until size 1. For n=8, the total calls are 2*n - 1 = 15.

🧠 Conceptual

intermediate

Master Theorem Case Identification

Given the recurrence relation T(n) = 3T(n/4) + n, which case of the Master Theorem applies?

A Case 2: f(n) = Θ(n^log_b(a) * log^k(n)), solution T(n) = Θ(n^log_b(a) * log^{k+1}(n))

B Case 1: f(n) = O(n^log_b(a)-ε), solution T(n) = Θ(n^log_b(a))

C Case 3: f(n) = Ω(n^log_b(a)+ε), solution T(n) = Θ(f(n))

D None of the above

💡 Hint

Calculate log base b of a and compare f(n) with n^{log_b(a)}.

Here, a=3, b=4, so log_b(a) = log_4(3) ≈ 0.792. Since f(n) = n = n^1, and 1 > 0.792, f(n) grows faster than n^{log_b(a)}. But since f(n) is polynomially larger, it fits Case 3.

🔧 Debug

advanced

Identify the Error in Recurrence Implementation

This Java method attempts to compute the time complexity T(n) = 2T(n/2) + n using recursion and returns the total cost. What error will occur when running computeTime(0)?

DSA Java

public class Recurrence {
    public static int computeTime(int n) {
        if (n == 1) {
            return 1;
        }
        return 2 * computeTime(n / 2) + n;
    }

    public static void main(String[] args) {
        System.out.println(computeTime(0));
    }
}

A ArithmeticException due to division by zero

B Returns 0 without error

C Compilation error

D StackOverflowError due to infinite recursion

💡 Hint

Check the base case and what happens when n=0.

When n=0, the base case (n==1) is never reached. The method calls computeTime(0/2) = computeTime(0) infinitely, causing StackOverflowError.

🚀 Application

advanced

Calculate Number of Subproblems in Divide and Conquer

A divide and conquer algorithm splits a problem of size n into 5 subproblems each of size n/3, and does linear work outside the recursive calls. How many subproblems are created at the third level of recursion?

A 125

B 15

C 27

D 81

💡 Hint

Number of subproblems at level k is a^k where a is the number of subproblems per call.

At each level, the number of subproblems multiplies by 5. At level 3, total subproblems = 5^3 = 125.

🧠 Conceptual

expert

Exact Solution of Recurrence Using Recursion Tree

Given the recurrence T(n) = 4T(n/2) + n^2, what is the exact asymptotic solution for T(n)?

A Θ(n^2 log n)

B Θ(n^2)

C Θ(n^3)

D Θ(n^{log_2 4})

💡 Hint

Use Master Theorem: a=4, b=2, f(n)=n^2, compare f(n) with n^{log_b a}.

Here, a=4, b=2, so log_b(a) = log_2(4) = 2. f(n) = n^2 matches n^{log_b(a)}. This is Master Theorem Case 2, so T(n) = Θ(n^2 log n).

Practice

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

easy

A. Store all intermediate results to avoid repeated calculations

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

C. Use a single loop to iterate through all elements

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

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 subproblems

   After solving smaller parts, 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 D

4. Quick Check:

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

Hint: Divide big problem 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) = T(n/2) + O(1)

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

C. T(n) = T(n-1) + 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, 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)
• Ignoring the merge cost O(n)
• Confusing with quicksort recurrence

3. What is the output of the following Java method that uses divide and conquer?

public static int sumArray(int[] arr, int start, int end) {
    if (start == end) return arr[start];
    int mid = (start + end) / 2;
    return sumArray(arr, start, mid) + sumArray(arr, mid + 1, end);
}

public static void main(String[] args) {
    int[] nums = {2, 4, 6, 8};
    System.out.println(sumArray(nums, 0, nums.length - 1));
}

medium

A. 20

B. 10

C. 0

D. Compilation error

Solution

1. Step 1: Understand the method's logic

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

2. Step 2: Calculate sum of array elements

   Sum of {2,4,6,8} = 2+4+6+8 = 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Sum of array = 20 [OK]

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

Common Mistakes:

• Confusing mid calculation causing infinite recursion
• Forgetting to include last element
• Assuming method returns partial sums

4. Identify the error in the following recurrence relation code snippet for a divide and conquer algorithm:

int divideConquer(int n) {
    if (n == 1) return 1;
    int result = divideConquer(n / 2) + divideConquer(n / 2);
    return result + n;
}

medium

A. The base case should be n == 0 instead of n == 1

B. No error, code is correct

C. The function calls divideConquer(n / 2) twice causing redundant work

D. The addition of n should be outside the return statement

Solution

1. Step 1: Analyze recursive calls

   The function calls divideConquer(n/2) twice separately, repeating work.

2. Step 2: Identify redundancy

   Calling the same function twice with same argument duplicates effort, inefficient.

3. Final Answer:

   The function calls divideConquer(n / 2) twice causing redundant work -> Option C

4. Quick Check:

   Repeated calls with same input cause redundancy [OK]

Hint: Avoid duplicate recursive calls with same parameters [OK]

Common Mistakes:

• Thinking base case is wrong
• Ignoring duplicate calls
• Assuming addition placement causes error

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

hard

A. O(n log n)

B. O(n^2)

C. O(n)

D. O(log n)

Solution

1. Step 1: Identify parameters for Master Theorem

   Here, a=3, b=3, f(n)=n.

2. Step 2: Calculate log_b(a)

   log_3(3) = 1.

3. Step 3: 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)}).

4. Step 4: Apply Master Theorem case 2

   Time complexity T(n) = Theta(n log n).

5. Final Answer:

   O(n log n) -> Option A

6. Quick Check:

   Master theorem case 2 = n log n [OK]

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

Common Mistakes:

• Confusing log base or value
• Choosing wrong Master theorem case
• Ignoring the additive term n