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.