Merge Sort as Divide and Conquer in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the 'Divide and Conquer' approach in Merge Sort?

Divide the array into two halves, sort each half recursively, and then merge the two sorted halves into one sorted array.

beginner

In Merge Sort, what happens during the 'merge' step?

Two sorted subarrays are combined into a single sorted array by comparing elements one by one and placing the smaller element first.

intermediate

Why is Merge Sort considered efficient for large datasets?

Because it consistently divides the problem into smaller parts and merges them in O(n log n) time, which is faster than simple sorting methods like bubble sort for large data.

beginner

What is the base case in the recursive Merge Sort algorithm?

When the array has one or zero elements, it is already sorted, so the recursion stops.

intermediate

How does Merge Sort handle the original array during sorting?

It does not sort in place; it uses extra space to merge sorted subarrays into a new array before copying back.

What is the time complexity of Merge Sort in the average case?

A O(n log n)

B O(n^2)

C O(log n)

D O(n)

Merge Sort divides the array and merges in O(n log n) time consistently.

During Merge Sort, what is the size of the subarrays when the recursion stops?

A One or zero elements

B Two elements

C Half the original array

D The entire array

The base case is when the subarray has one or zero elements, which are already sorted.

Which step in Merge Sort combines two sorted arrays into one sorted array?

A Split step

B Divide step

C Merge step

D Sort step

The merge step combines two sorted arrays into one sorted array.

What kind of algorithm is Merge Sort?

A Backtracking

B Greedy

C Dynamic Programming

D Divide and Conquer

Merge Sort uses the divide and conquer strategy.

Does Merge Sort sort the array in place without extra space?

A Yes, it sorts in place

B No, it requires extra space for merging

C Only for arrays smaller than 10 elements

D Only if the array is already sorted

Merge Sort requires extra space to merge sorted subarrays.

Explain how Merge Sort uses the divide and conquer approach to sort an array.

Think about breaking the problem into smaller parts and then combining results.

Describe the merge step in Merge Sort and why it is important.

Focus on how two sorted parts become one sorted whole.

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