Merge Sort as Divide and Conquer in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle index for dividing the array.

DSA C

int mid = (left [1] right) / 2;

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition

Dividing before adding

The middle index is found by adding left and right, then dividing by 2.

2fill in blank

medium

Complete the recursive call to sort the left half of the array.

DSA C

mergeSort(arr, [1], mid);

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

Aleft

B0

Cright

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of left

Using right instead of left

The left boundary is passed as the start index for sorting the left half.

3fill in blank

hard

Fix the error in the merge function call to merge two halves correctly.

DSA C

merge(arr, left, [1], right);

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

Aright - 1

Bmid - 1

Cleft + 1

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 instead of mid + 1

Using left + 1 incorrectly

The middle index plus one is the start of the right half for merging.

4fill in blank

hard

Fill both blanks to complete the loop copying elements back to the original array.

DSA C

for (int k = [1]; k <= [2]; k++) {
    arr[k] = temp[k];
}

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

Aleft

Bmid

Cright

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0 instead of left

Ending at mid instead of right

The loop copies elements from left to right indices back to the original array.

5fill in blank

hard

Fill all three blanks to complete the merge function's main merging loop.

DSA C

while (i <= [1] && j <= [2]) {
    if (arr[i] <= arr[j]) {
        temp[k++] = arr[[3]++];
    } else {
        temp[k++] = arr[j++];
    }
}

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

Amid

Bright

Ci

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong boundary values

Incrementing wrong index

The loop runs while i is less or equal to mid and j less or equal to right. The element at i is copied and i is incremented.

Practice

1. What is the main idea behind the Merge Sort algorithm?

easy

A. Divide the array into halves, sort each half, then merge them

B. Sort the array by repeatedly swapping adjacent elements

C. Select a pivot and partition the array around it

D. Insert elements one by one into a sorted sublist

Solution

1. Step 1: Understand the divide and conquer approach

   Merge Sort splits the array into two halves recursively until single elements remain.

2. Step 2: Merge sorted halves

   After splitting, it merges the sorted halves back to form a sorted array.

3. Final Answer:

   Divide the array into halves, sort each half, then merge them -> Option A

4. Quick Check:

   Merge Sort = Divide and Conquer [OK]

Hint: Remember: split, sort halves, then merge [OK]

Common Mistakes:

• Confusing Merge Sort with Bubble Sort
• Thinking Merge Sort uses a pivot like Quick Sort
• Believing Merge Sort sorts in place without merging

2. Which of the following is the correct base case condition in a recursive Merge Sort function in C?

easy

A. if (start >= end) return;

B. if (start > end) return;

C. if (start == end) return;

D. if (start != end) return;

Solution

1. Step 1: Identify when recursion should stop

   Merge Sort stops splitting when the subarray has one or zero elements, i.e., start >= end.

2. Step 2: Choose the correct condition

   The base case is when start >= end, so return without further splitting.

3. Final Answer:

   if (start >= end) return; -> Option A

4. Quick Check:

   Base case = start >= end [OK]

Hint: Stop recursion when start >= end [OK]

Common Mistakes:

• Using == instead of >= risks issues if start > end
• Using > misses the single element case
• Using != returns too early, breaking recursion

3. What will be the output of the following C code snippet after calling mergeSort(arr, 0, 4); on int arr[] = {5, 3, 8, 4, 2};?

void merge(int arr[], int l, int m, int r) {
    int n1 = m - l + 1, n2 = r - m;
    int L[n1], R[n2];
    for (int i = 0; i < n1; i++) L[i] = arr[l + i];
    for (int j = 0; j < n2; j++) R[j] = arr[m + 1 + j];
    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) arr[k++] = L[i++];
        else arr[k++] = R[j++];
    }
    while (i < n1) arr[k++] = L[i++];
    while (j < n2) arr[k++] = R[j++];
}

void mergeSort(int arr[], int l, int r) {
    if (l < r) {
        int m = l + (r - l) / 2;
        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);
        merge(arr, l, m, r);
    }
}

int main() {
    int arr[] = {5, 3, 8, 4, 2};
    mergeSort(arr, 0, 4);
    for (int i = 0; i < 5; i++) printf("%d ", arr[i]);
    return 0;
}

medium

A. 5 3 8 4 2

B. 8 5 4 3 2

C. 2 3 4 5 8

D. 2 4 3 5 8

Solution

1. Step 1: Understand mergeSort splits and merges

   The array is recursively split into halves and merged in sorted order.

2. Step 2: Trace the final sorted array

   After full recursion and merging, the array becomes sorted: 2 3 4 5 8.

3. Final Answer:

   2 3 4 5 8 -> Option C

4. Quick Check:

   Sorted array output = 2 3 4 5 8 [OK]

Hint: Merge Sort always outputs sorted array [OK]

Common Mistakes:

• Thinking mergeSort does not change original array
• Confusing mergeSort with reverse sorting
• Ignoring the merge step results in unsorted output

4. Identify the error in the following merge function snippet used in Merge Sort:

void merge(int arr[], int l, int m, int r) {
    int n1 = m - l + 1, n2 = r - m;
    int L[n1], R[n2];
    for (int i = 0; i <= n1; i++) L[i] = arr[l + i];
    for (int j = 0; j <= n2; j++) R[j] = arr[m + 1 + j];
    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) arr[k++] = L[i++];
        else arr[k++] = R[j++];
    }
    while (i < n1) arr[k++] = L[i++];
    while (j < n2) arr[k++] = R[j++];
}

medium

A. The merge function should not use temporary arrays

B. The loops copying to L and R arrays use <= instead of < causing out-of-bounds access

C. The size of arrays L and R is calculated incorrectly

D. The merge condition should be L[i] < R[j] instead of <= to avoid duplicates

Solution

1. Step 1: Check array copying loops

   The loops use i <= n1 and j <= n2 which causes index to go out of bounds since arrays are 0-based.

2. Step 2: Correct loop condition

   Loops should use i < n1 and j < n2 to avoid accessing invalid memory.

3. Final Answer:

   The loops copying to L and R arrays use <= instead of < causing out-of-bounds access -> Option B

4. Quick Check:

   Loop bounds must be less than array size [OK]

Hint: Use < not <= in loops for array indexing [OK]

Common Mistakes:

• Using <= causes buffer overflow errors
• Confusing array size with last index
• Ignoring array indexing starting at 0

5. You want to sort a linked list using Merge Sort. Which of the following is the best approach to split the list into two halves efficiently?

hard

A. Convert the list to an array, sort it, then rebuild the list

B. Count the total nodes first, then traverse again to the middle node

C. Split the list by breaking after the first node repeatedly

D. Use two pointers: move one pointer twice as fast as the other to find the middle

Solution

1. Step 1: Understand splitting linked list for Merge Sort

   Using two pointers (slow and fast), where fast moves twice as fast, helps find the middle in one pass.

2. Step 2: Compare other methods

   Counting nodes requires two passes; breaking after first node is incorrect; converting to array is inefficient.

3. Final Answer:

   Use two pointers: move one pointer twice as fast as the other to find the middle -> Option D

4. Quick Check:

   Two-pointer technique = efficient split [OK]

Hint: Use slow-fast pointer to find middle in one pass [OK]

Common Mistakes:

• Counting nodes wastes time with two traversals
• Breaking after first node does not split evenly
• Converting to array loses linked list benefits