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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

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

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

Amid

Blow

Chigh

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high.

Using length instead of high index.

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.

DSA Python

result_left = divide_and_conquer(arr, low, [1])

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

Amid + 1

Bhigh

Clow

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using high instead of mid.

Using mid + 1 which is for right half.

The left half is from low to mid.

3fill in blank

hard

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

DSA Python

if low [1] high:
    return arr[low]

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

A>=

B>

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or > without equality.

Using wrong comparison direction.

The base case occurs when low is equal to or greater than high, meaning one element or invalid segment.

4fill in blank

hard

Fill both blanks to correctly combine results from left and right halves.

DSA Python

return [1](result_left, result_right) if result_left [2] result_right else result_right

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Amax

B>

C<

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max.

Using less than operator incorrectly.

We use max to combine results and compare if result_left is greater than result_right.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension for recurrence relations.

DSA Python

recurrence = { [1]: [2] for n in range(1, 6) if n [3] 3 }

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An

B2**n

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong keys or values.

Using incorrect comparison operator.

The dictionary keys are n, values are 2**n, and we include only keys greater than 3.

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 solve the problem iteratively

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 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 A

4. Quick Check:

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

Hint: Divide big problem into smaller parts, solve, 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 syntax for the recurrence relation of Merge Sort?

easy

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

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

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

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

Solution

1. Step 1: Recall Merge Sort recurrence

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

2. Step 2: Write recurrence relation

   It calls itself twice on half size (2T(n/2)) plus O(n) for merging.

3. Final Answer:

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

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 as O(n^2)
• Choosing wrong number of recursive calls

3. What is the output of this Python function that uses divide and conquer to compute sum of a list?

def sum_list(arr):
    if len(arr) == 1:
        return arr[0]
    mid = len(arr) // 2
    left_sum = sum_list(arr[:mid])
    right_sum = sum_list(arr[mid:])
    return left_sum + right_sum

print(sum_list([2, 4, 6, 8]))

medium

A. 16

B. 20

C. Error

D. 10

Solution

1. Step 1: Trace recursive calls

   List splits into [2,4] and [6,8], then further into single elements.

2. Step 2: Calculate sums bottom-up

   Sum of [2,4] = 6, sum of [6,8] = 14, total = 6 + 14 = 20.

3. Final Answer:

   20 -> Option B

4. Quick Check:

   2+4+6+8 = 20 [OK]

Hint: Sum divides list, sums parts, then adds [OK]

Common Mistakes:

• Confusing mid index slicing
• Returning partial sums incorrectly
• Forgetting base case for single element

4. Find the error in this divide and conquer function to find maximum element in a list:

def max_element(arr):
    if len(arr) == 1:
        return arr[0]
    mid = len(arr) // 2
    left_max = max_element(arr[:mid])
    right_max = max_element(arr[mid:])
    if left_max > right_max:
        return left_max
    else:
        return arr[right_max]

medium

A. Function should use iteration instead of recursion

B. Base case missing for empty list

C. Mid calculation should be mid = len(arr) // 3

D. Incorrect return statement: should return right_max, not arr[right_max]

Solution

1. Step 1: Check return statements

   Left max returns value correctly, right max returns index incorrectly by arr[right_max].

2. Step 2: Fix return for right max comparison

   Should return right_max directly, not arr[right_max], because right_max is already a value.

3. Final Answer:

   Incorrect return statement: should return right_max, not arr[right_max] -> Option D

4. Quick Check:

   Return max values, not index or wrong element [OK]

Hint: Return max values, not arr[value] [OK]

Common Mistakes:

• Returning element by index instead of value
• Wrong slicing causing infinite recursion
• Ignoring base case for single element

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

hard

A. O(n)

B. O(n^2)

C. O(n log 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 n^log_b(a)

   log_3(3) = 1, so n^1 = n.

3. Step 3: Compare f(n) with n^log_b(a)

   f(n) = n matches n^log_b(a) = n, so case 2 applies.

4. Step 4: Apply Master Theorem case 2

   Time complexity is O(n log n).

5. Final Answer:

   O(n log n) -> Option C

6. Quick Check:

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

Hint: Compare f(n) to n^log_b(a) for Master Theorem [OK]

Common Mistakes:

• Confusing log base or value
• Choosing wrong Master Theorem case
• Ignoring the additive term f(n)