Merge Sort Algorithm in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Merge Sort on a small array

What is the output of the following C++ code after performing merge sort on the array?

DSA C++

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];
            i++;
        } else {
            arr[k] = R[j];
            j++;
        }
        k++;
    }
    while (i < n1) {
        arr[k] = L[i];
        i++;
        k++;
    }
    while (j < n2) {
        arr[k] = R[j];
        j++;
        k++;
    }
}

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, 2, 9, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    mergeSort(arr, 0, n - 1);
    for (int i = 0; i < n; i++) {
        std::cout << arr[i] << " ";
    }
    return 0;
}

A 1 2 5 9

B 9 5 2 1

C 2 1 5 9

D 1 5 2 9

💡 Hint

Remember that merge sort divides the array and merges sorted halves.

Merge sort sorts the array in ascending order by dividing and merging. The sorted array is 1 2 5 9.

🧠 Conceptual

intermediate

Time Complexity of Merge Sort

What is the time complexity of merge sort in the average and worst cases?

A O(log n)

B O(n log n)

C O(n^2)

D O(n)

💡 Hint

Think about how many times the array is divided and merged.

Merge sort divides the array into halves log n times and merges n elements each time, resulting in O(n log n).

🔧 Debug

advanced

Identify the bug in this merge function

What error will this merge function cause when run?

DSA C++

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];
            i++;
        } else {
            arr[k] = R[j];
            j++;
        }
        k++;
    }
    while (i < n1) {
        arr[k] = L[i];
        i++;
        k++;
    }
    while (j < n2) {
        arr[k] = R[j];
        j++;
        k++;
    }
}

A Array index out of bounds error

B No error, runs correctly

C Infinite loop

D Compilation error due to missing return

💡 Hint

Check the for loop conditions carefully.

The loop for copying L uses i <= n1 which accesses one element beyond array size causing out of bounds error.

❓ Predict Output

advanced

Output after sorting a reversed array

What is the output of the following code after merge sorting the array?

DSA C++

int main() {
    int arr[] = {10, 9, 8, 7, 6};
    int n = sizeof(arr) / sizeof(arr[0]);
    mergeSort(arr, 0, n - 1);
    for (int i = 0; i < n; i++) {
        std::cout << arr[i] << " ";
    }
    return 0;
}

A 6 8 7 9 10

B 10 9 8 7 6

C 7 6 8 9 10

D 6 7 8 9 10

💡 Hint

Merge sort sorts in ascending order regardless of initial order.

The reversed array is sorted into ascending order: 6 7 8 9 10.

🧠 Conceptual

expert

Space complexity of merge sort

What is the space complexity of merge sort and why?

A O(log n) because of recursive call stack only

B O(1) because it sorts in place without extra space

C O(n) because it requires temporary arrays for merging

D O(n log n) because it copies the array multiple times

💡 Hint

Think about the extra arrays used during merging.

Merge sort uses temporary arrays to merge halves, requiring O(n) extra space.

Practice

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

easy

A. Select the smallest element and place it at the beginning

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

C. Sort the array by repeatedly swapping adjacent elements

D. Build a heap and extract elements in sorted order

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

   It then merges these sorted halves to produce a fully sorted array.

3. Final Answer:

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

4. Quick Check:

   Merge Sort = Divide, Sort, Merge [OK]

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

Common Mistakes:

• Confusing merge sort with bubble sort
• Thinking merge sort swaps adjacent elements
• Mixing merge sort with selection sort

2. Which of the following is the correct way to declare the merge function header in C++ for merge sort?

easy

A. int mergeSort(int arr[], int left, int right);

B. int merge(int arr[], int left, int mid, int right);

C. void mergeSort(int arr[], int left, int right);

D. void merge(int arr[], int left, int mid, int right);

Solution

1. Step 1: Identify the merge function role

   The merge function combines two sorted parts; it does not return a value, so it should be void.

2. Step 2: Check parameter list

   It takes the array and three indices: left, mid, and right to merge subarrays.

3. Final Answer:

   void merge(int arr[], int left, int mid, int right); -> Option D

4. Quick Check:

   Merge function = void with arr and indices [OK]

Hint: Merge combines parts, returns nothing [OK]

Common Mistakes:

• Using int return type for merge
• Confusing merge with mergeSort function
• Wrong parameter order or missing parameters

3. What will be the output of the following code snippet after applying merge sort on the array?

int arr[] = {5, 2, 9, 1};
mergeSort(arr, 0, 3);
for(int i = 0; i < 4; i++) std::cout << arr[i] << " ";

medium

A. 5 2 9 1

B. 9 5 2 1

C. 1 2 5 9

D. 2 1 5 9

Solution

1. Step 1: Apply merge sort on the array

   The array {5, 2, 9, 1} is split and sorted recursively to {1, 2, 5, 9}.

2. Step 2: Print the sorted array

   The for loop prints elements in sorted order: 1 2 5 9.

3. Final Answer:

   1 2 5 9 -> Option C

4. Quick Check:

   Sorted array = 1 2 5 9 [OK]

Hint: Merge sort outputs sorted array ascending [OK]

Common Mistakes:

• Printing original array without sorting
• Confusing merge sort with reverse sort
• Off-by-one errors in indices

4. Identify the error in the following merge sort code snippet:

void merge(int arr[], int left, int mid, int right) {
  int n1 = mid - left;
  int n2 = right - mid;
  int L[n1], R[n2];
  for(int i = 0; i < n1; i++) L[i] = arr[left + i];
  for(int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];
  // merging logic...
}

medium

A. Incorrect calculation of n1 and n2 sizes

B. Wrong indexing in copying elements to L array

C. Using variable-length arrays which is invalid in C++

D. Missing return statement in void function

Solution

1. Step 1: Check size calculations

   n1 = mid - left is missing +1 (should be mid - left + 1) to include the mid element; n2 = right - mid is correct.

2. Step 2: Check copying to L array

   Copying starts at left + i for i < n1, which goes up to mid-1, missing arr[mid].

3. Step 3: Identify the main error

   The error causes wrong indexing/range when copying elements to L array.

4. Final Answer:

   Incorrect calculation of n1 and n2 sizes -> Option A

5. Quick Check:

   Subarray size calculation off by one [OK]

Hint: Check array copy indices carefully [OK]

Common Mistakes:

• Off-by-one errors in subarray sizes
• Using variable-length arrays in C++
• Ignoring void function return type

5. Given an array with many duplicate elements, how can merge sort be adapted to maintain the relative order of equal elements (stable sort)?

hard

A. Always pick the element from the left subarray first when merging equals

B. Always pick the element from the right subarray first when merging equals

C. Swap equal elements randomly during merge

D. Use quicksort instead of merge sort

Solution

1. Step 1: Understand stability in sorting

   A stable sort keeps equal elements in their original relative order.

2. Step 2: Apply stability in merge step

   When merging, if elements are equal, always choose the element from the left subarray first to keep order.

3. Final Answer:

   Always pick the element from the left subarray first when merging equals -> Option A

4. Quick Check:

   Stable merge = left element first on equals [OK]

Hint: Pick left element first on ties to keep order [OK]

Common Mistakes:

• Picking right element first breaks stability
• Random swaps destroy order
• Confusing merge sort with quicksort stability