Why Divide and Conquer and What It Gives You in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to divide the problem into two halves.

DSA C

int mid = (start + end) [1] 2;

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A*

B/

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Using addition or subtraction instead of division.

We use division by 2 to find the middle point between start and end.

2fill in blank

medium

Complete the code to recursively solve the left half of the problem.

DSA C

divideAndConquer(array, start, [1]);

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Amid

Bend

Cstart

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using end instead of mid for the left half.

Using mid + 1 which belongs to the right half.

We pass mid as the end index for the left half.

3fill in blank

hard

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

DSA C

if (start [1] end) return;

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A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using equality instead of greater than.

Using less than which causes infinite recursion.

The recursion stops when start is greater than end, meaning no elements to process.

4fill in blank

hard

Fill both blanks to recursively solve left and right halves.

DSA C

divideAndConquer(array, start, [1]);
divideAndConquer(array, [2], end);

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Amid

Bstart

Cmid + 1

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using start or end incorrectly for halves.

Mixing up mid and mid + 1.

The left half ends at mid, the right half starts at mid + 1.

5fill in blank

hard

Fill all three blanks to merge two sorted halves correctly.

DSA C

int i = start, j = [1], k = 0;
while (i <= [2] && j <= [3]) {
    if (array[i] < array[j]) {
        temp[k++] = array[i++];
    } else {
        temp[k++] = array[j++];
    }
}

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

Amid + 1

Bmid

Cend

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Starting j at mid instead of mid + 1.

Using wrong end points for i or j.

We start j at mid + 1, i goes up to mid, and j goes up to end to merge halves.

Practice

1. What is the main advantage of using the Divide and Conquer approach in algorithms?

easy

A. It uses more memory to store all intermediate results

B. It always runs in constant time

C. It avoids recursion completely

D. It breaks a problem into smaller subproblems to solve them independently

Solution

1. Step 1: Understand the divide and conquer method

   Divide and conquer splits a problem into smaller parts that are easier to solve.

2. Step 2: Identify the main benefit

   By solving smaller subproblems independently, the overall problem becomes simpler and often faster to solve.

3. Final Answer:

   It breaks a problem into smaller subproblems to solve them independently -> Option D

4. Quick Check:

   Divide and conquer = break into smaller parts [OK]

Hint: Divide problems into smaller parts to solve easily [OK]

Common Mistakes:

• Thinking divide and conquer avoids recursion
• Believing it always uses constant time
• Confusing it with dynamic programming

2. Which of the following is the correct syntax to recursively call a function divideAndConquer in C?

easy

A. divideAndConquer(); // recursive call

B. divideAndConquer;

C. divideAndConquer()

D. call divideAndConquer();

Solution

1. Step 1: Identify correct function call syntax in C

   In C, a function is called by writing its name followed by parentheses and semicolon.

2. Step 2: Check which option matches this syntax

   divideAndConquer(); // recursive call shows the correct syntax with a comment explaining it's a recursive call.

3. Final Answer:

   divideAndConquer(); // recursive call -> Option A

4. Quick Check:

   Function call syntax = name() [OK]

Hint: Function calls need parentheses and semicolon [OK]

Common Mistakes:

• Omitting parentheses in function call
• Using 'call' keyword which is invalid in C
• Missing semicolon at end

3. What will be the output of the following C code snippet?

void printNumbers(int n) {
  if (n == 0) return;
  printNumbers(n / 2);
  printf("%d", n % 2);
}

int main() {
  printNumbers(6);
  return 0;
}

medium

A. 011

B. 110

C. 101

D. 100

Solution

1. Step 1: Trace recursive calls with n=6

   Calls: printNumbers(6) calls printNumbers(3), which calls printNumbers(1), which calls printNumbers(0) and returns.

2. Step 2: Print n % 2 on return

   Return from printNumbers(0) prints nothing.
   printNumbers(1) prints 1 (1 % 2 = 1).
   printNumbers(3) prints 1 (3 % 2 = 1).
   printNumbers(6) prints 0 (6 % 2 = 0).

3. Final Answer:

   110 -> Option B

4. Quick Check:

   Binary of 6 = 110 [OK]

Hint: Divide n by 2 recursively to print binary [OK]

Common Mistakes:

• Printing before recursive call instead of after
• Confusing n/2 with n%2
• Not handling base case correctly

4. Identify the error in the following divide and conquer function for finding the maximum element in an array:

int findMax(int arr[], int start, int end) {
  if (start == end) return arr[start];
  int mid = (start + end) / 2;
  int leftMax = findMax(arr, start, mid);
  int rightMax = findMax(arr, mid + 1, end);
  if (leftMax > rightMax) return leftMax;
  else return rightMax;
}

// Called as findMax(arr, 0, n);

medium

A. The function call should be findMax(arr, 0, n-1)

B. The mid calculation causes infinite recursion

C. The base case condition is incorrect

D. The comparison between leftMax and rightMax is wrong

Solution

1. Step 1: Check function call parameters

   Array indices go from 0 to n-1, so calling with end = n is out of bounds.

2. Step 2: Correct the call to avoid errors

   Call should be findMax(arr, 0, n-1) to cover full array correctly without overflow.

3. Final Answer:

   The function call should be findMax(arr, 0, n-1) -> Option A

4. Quick Check:

   Array indices end at n-1 [OK]

Hint: Use n-1 as last index in array calls [OK]

Common Mistakes:

• Using n as last index causing out-of-bounds
• Misunderstanding base case for single element
• Incorrect mid calculation causing infinite recursion

5. You want to sort an array using divide and conquer. Which of these steps correctly describes what happens in merge sort?

hard

A. Merge the array first, then divide it into smaller parts

B. Divide array into halves, sort only the first half, then append the second half

C. Divide array into halves, sort each half recursively, then merge sorted halves

D. Sort the entire array first, then divide it into halves

Solution

1. Step 1: Understand merge sort steps

   Merge sort divides the array into two halves, sorts each half recursively, then merges the sorted halves.

2. Step 2: Identify correct description

   Divide array into halves, sort each half recursively, then merge sorted halves correctly describes this process, while others do not follow the divide and conquer method properly.

3. Final Answer:

   Divide array into halves, sort each half recursively, then merge sorted halves -> Option C

4. Quick Check:

   Merge sort = divide, conquer, merge [OK]

Hint: Divide, sort halves, then merge sorted parts [OK]

Common Mistakes:

• Sorting only one half and ignoring the other
• Merging before sorting
• Sorting entire array before dividing