Merge Sort as Divide and Conquer in DSA C - Time & Space Complexity

Time Complexity: Merge Sort as Divide and Conquer

O(n log n)

Understanding Time Complexity

We want to understand how the time needed by merge sort grows as the list gets bigger.

Specifically, how does splitting and merging affect the total work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void merge(int arr[], int l, int m, int r) {
    int n1 = m - l + 1;
    int 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);
    }
}

This code splits the array into halves, sorts each half recursively, then merges them back.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The merge function compares and copies elements to combine two sorted halves.
• How many times: MergeSort calls itself twice on halves, creating a tree of calls with depth about log n.
• Each merge step processes all elements in the current segment once.

How Execution Grows With Input

Each level of splitting does work proportional to n, and there are about log n levels.

Input Size (n)	Approx. Operations
10	About 10 * 4 = 40
100	About 100 * 7 = 700
1000	About 1000 * 10 = 10,000

Pattern observation: Operations grow a bit more than linearly, roughly n times log n.

Final Time Complexity

Time Complexity: O(n log n)

This means if the list doubles in size, the work grows a bit more than double, because of the splitting and merging steps.

Common Mistake

[X] Wrong: "Merge sort is just like doing a simple loop over the array, so it is O(n)."

[OK] Correct: Merge sort splits the array many times and merges at each level, so the total work adds up to more than just one pass.

Interview Connect

Understanding merge sort's time complexity shows you how divide and conquer helps solve problems efficiently, a key skill in many coding challenges.

Self-Check

"What if we changed merge sort to split the array into three parts instead of two? How would the time complexity change?"

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