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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

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

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

Alength

Blow

Cmid

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high

Using length which is not defined here

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

2fill in blank

medium

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

DSA Java

mergeSort(arr, low, [1]);

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Amid

Bmid + 1

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 which is for right half

Using high which is the end of the whole array

The left half is from low to mid.

3fill in blank

hard

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

DSA Java

if (low [1] high) return;

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A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or <= which causes infinite recursion

Using > without equality misses single element case

The recursion stops when low is greater than or equal to high, meaning one or no elements.

4fill in blank

hard

Fill both blanks to complete the recurrence relation for merge sort.

DSA Java

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

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A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of division

Using multiplication instead of addition

The recurrence relation splits the problem into two halves of size n/2 and adds linear time cn for merging.

5fill in blank

hard

Fill all three blanks to complete the Master Theorem case for merge sort.

DSA Java

If a = 2, b = 2, and f(n) = cn, then T(n) = [1](n \lg n) where n^[2] [3] f(n).

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ATheta

Blog_b a

C=

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using inequality signs instead of equals

Confusing log base and exponent

According to the Master Theorem case 2, when f(n) = \Theta(n^{log_b a}), the solution is \Theta(n^{log_b a} \lg 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