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++];
    }
}

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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.