Exercises · Introduction to C++¶

Exercises · Chapter 6 · each solution opens with a click · Chapter 6: Introduction to C++ · Exercises, volume 2 (PDF)

In the exercises of this chapter the programs are written, compiled and executed with the conventions of the chapter «Introduction to C++», summarized below.

Code

The program of an exercise is saved in a file <name>.cpp, compiled and executed with the commands
```
g++ -Wall -std=c++17 -o <name> <name>.cpp
./<name>
```
The test instances are fixed-size arrays declared in main, for example int arr1[] = {9, 12, 3, 16, 19, 11, 12};, and are passed to the functions through the pointer to their first element (int *a) together with their size n.
The elements of an array of size \(n\) are a[0], a[1], …, a[n-1]: the entry \(a[i]\) of an instance written as \(a = [\dots]\) in the text corresponds to a[i-1] in the code.
The results that a function has to give back besides its return value are passed by reference (int& x, double& x).
When an array is printed, every value is followed by a space, so the printed lines end with a space.

Exercise 6.1: Introduction to {\tt C++} (E1)

Given an array \(a\) of \(n \ge 1\) integer numbers, we consider the problem of finding the maximum value in the array.

Questions

Write a complete C++ program that finds the maximum value in an array \(a\) of \(n\) integer numbers. The program should:
- Define a function int ArrayMaxValue(int n, int *a) that implements the algorithm
- In the main() function, create test arrays and call the function
- Print the results for each test case
Execute the program on the following instances:
- Instance 1: \(a = [9, 12, 3, 16, 19, 11, 12]\) with \(n=7\)
- Instance 2: \(a = [5, 2, 8, 1]\) with \(n=4\)
- Instance 3: \(a = [42]\) with \(n=1\)
Show the complete compilation commands and output for all test cases.
Modify the program to create a new function int ArrayMaxValueWithIndex(int n, int *a, int& idx) that returns both the maximum value and stores its index in the reference parameter idx. Execute this modified version and show the output.

Solution

Complete C++ program:

#include <iostream>
using namespace std;

// Function to find maximum value in array
int ArrayMaxValue(int n, int *a) 
{
    int max = a[0];  // Initialize with first element

    for (int i = 1; i < n; i++) 
    {
        if (a[i] > max) 
        {
            max = a[i];  // Update maximum
        }
    }

    return max;
}

int main() 
{
    // Test case 1
    int arr1[] = {9, 12, 3, 16, 19, 11, 12};
    int n1 = 7;

    cout << "Test 1 - Array: ";
    for (int i = 0; i < n1; i++) 
    {
        cout << arr1[i] << " ";
    }
    cout << endl;

    int max1 = ArrayMaxValue(n1, arr1);
    cout << "Maximum value: " << max1 << endl;
    cout << endl;

    // Test case 2
    int arr2[] = {5, 2, 8, 1};
    int n2 = 4;

    cout << "Test 2 - Array: ";
    for (int i = 0; i < n2; i++) 
    {
        cout << arr2[i] << " ";
    }
    cout << endl;

    int max2 = ArrayMaxValue(n2, arr2);
    cout << "Maximum value: " << max2 << endl;
    cout << endl;

    // Test case 3
    int arr3[] = {42};
    int n3 = 1;

    cout << "Test 3 - Array: ";
    for (int i = 0; i < n3; i++) 
    {
        cout << arr3[i] << " ";
    }
    cout << endl;

    int max3 = ArrayMaxValue(n3, arr3);
    cout << "Maximum value: " << max3 << endl;

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o max_value max_value.cpp

To run the program:

./max_value

Output:

Test 1 - Array: 9 12 3 16 19 11 12 
Maximum value: 19

Test 2 - Array: 5 2 8 1 
Maximum value: 8

Test 3 - Array: 42 
Maximum value: 42

Modified program with index:

#include <iostream>
using namespace std;

// Function to find maximum value and its index
int ArrayMaxValueWithIndex(int n, int *a, int& idx) 
{
    int max = a[0];
    idx = 0;

    for (int i = 1; i < n; i++) 
    {
        if (a[i] > max) 
        {
            max = a[i];
            idx = i;  // Store the index
        }
    }

    return max;
}

int main() 
{
    int arr[] = {9, 12, 3, 16, 19, 11, 12};
    int n = 7;
    int index;

    cout << "Array: ";
    for (int i = 0; i < n; i++) 
    {
        cout << arr[i] << " ";
    }
    cout << endl;

    int max = ArrayMaxValueWithIndex(n, arr, index);

    cout << "Maximum value: " << max << endl;
    cout << "Position (index): " << index << endl;

    return 0;
}

Output:

Array: 9 12 3 16 19 11 12 
Maximum value: 19
Position (index): 4

Note: The maximum value 19 is found at index 4 (the 5th element, since arrays are indexed from \(0\) in C++). If the maximum value appears more than once, the function stores the index of its first occurrence, because max and idx are updated only when a[i] > max (strict inequality).

Exercise 6.2: Introduction to {\tt C++} (E2)

Given an array \(a\) of \(n \ge 1\) integer numbers, we consider the problem of finding its minimum value.

Questions

Write a complete C++ program that finds the minimum value in an array \(a\) of \(n\) integer numbers. The program should:
- Define a function int ArrayMinValue(int n, int *a) that implements the algorithm
- In the main() function, create test arrays and call the function
- Print the results for each test case
Execute the program on the following instances:
- Instance 1: \(a = [9, 12, 3, 16, 19, 11, 12]\) with \(n=7\)
- Instance 2: \(a = [15, 7, 22, 4]\) with \(n=4\)
- Instance 3: \(a = [100]\) with \(n=1\)
Show the complete compilation commands and output for all test cases.
Modify the program to create a new function void ArrayMaxMin(int n, int *a, int& max, int& min) that finds both the maximum and minimum values in a single pass through the array. Execute this modified version and show the output.

Solution

Complete C++ program:

#include <iostream>
using namespace std;

// Function to find minimum value in array
int ArrayMinValue(int n, int *a) 
{
    int m = a[0];  // Initialize with first element

    for (int i = 1; i < n; i++) 
    {
        if (a[i] < m) 
        {
            m = a[i];  // Update minimum
        }
    }

    return m;
}

int main() 
{
    // Test case 1
    int arr1[] = {9, 12, 3, 16, 19, 11, 12};
    int n1 = 7;

    cout << "Test 1 - Array: ";
    for (int i = 0; i < n1; i++) 
    {
        cout << arr1[i] << " ";
    }
    cout << endl;

    int min1 = ArrayMinValue(n1, arr1);
    cout << "Minimum value: " << min1 << endl;
    cout << endl;

    // Test case 2
    int arr2[] = {15, 7, 22, 4};
    int n2 = 4;

    cout << "Test 2 - Array: ";
    for (int i = 0; i < n2; i++) 
    {
        cout << arr2[i] << " ";
    }
    cout << endl;

    int min2 = ArrayMinValue(n2, arr2);
    cout << "Minimum value: " << min2 << endl;
    cout << endl;

    // Test case 3
    int arr3[] = {100};
    int n3 = 1;

    cout << "Test 3 - Array: ";
    for (int i = 0; i < n3; i++) 
    {
        cout << arr3[i] << " ";
    }
    cout << endl;

    int min3 = ArrayMinValue(n3, arr3);
    cout << "Minimum value: " << min3 << endl;

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o min_value min_value.cpp

To run the program:

./min_value

Output:

Test 1 - Array: 9 12 3 16 19 11 12 
Minimum value: 3

Test 2 - Array: 15 7 22 4 
Minimum value: 4

Test 3 - Array: 100 
Minimum value: 100

Modified program to find both max and min:

#include <iostream>
using namespace std;

// Function to find both maximum and minimum values
void ArrayMaxMin(int n, int *a, int& max, int& min) 
{
    max = a[0];
    min = a[0];

    for (int i = 1; i < n; i++) 
    {
        if (a[i] > max) 
        {
            max = a[i];
        }
        if (a[i] < min) 
        {
            min = a[i];
        }
    }
}

int main() 
{
    int arr[] = {9, 12, 3, 16, 19, 11, 12};
    int n = 7;
    int maximum, minimum;

    cout << "Array: ";
    for (int i = 0; i < n; i++) 
    {
        cout << arr[i] << " ";
    }
    cout << endl;

    ArrayMaxMin(n, arr, maximum, minimum);

    cout << "Maximum value: " << maximum << endl;
    cout << "Minimum value: " << minimum << endl;

    return 0;
}

Output:

Array: 9 12 3 16 19 11 12 
Maximum value: 19
Minimum value: 3

Note: This implementation finds both the maximum and the minimum value in a single pass through the array. The number of comparisons is the same as with two separate functions, \(2(n-1)\), but the array is traversed once instead of twice.

Exercise 6.3: Introduction to {\tt C++} (E3)

Given an array \(a\) of \(n \ge 1\) integer numbers, we consider the problem of computing the sum and average of all its elements.

Questions

Write a complete C++ program that computes the sum and average of elements in an array \(a\) of \(n\) integer numbers. The program should:
- Define a function int ArraySum(int n, int *a) that computes the sum of array elements
- In the main() function, create test arrays and call the function
- Compute and print both the sum and average for each test case
Execute the program on the following instances:
- Instance 1: \(a = [5, 12, 3, 8]\) with \(n=4\)
- Instance 2: \(a = [10, 20, 30]\) with \(n=3\)
- Instance 3: \(a = [7]\) with \(n=1\)
Show the complete compilation commands and output for all test cases.
Modify the program to create a new function void ArraySumAndAverage(int n, int *a, int& sum, double& avg) that computes both the sum and average in one function. Execute this modified version and show the output.

Solution

Complete C++ program:

#include <iostream>
#include <iomanip>
using namespace std;

// Function to compute sum of array elements
int ArraySum(int n, int *a) 
{
    int S = 0;

    for (int i = 0; i < n; i++) 
    {
        S = S + a[i];
    }

    return S;
}

int main() 
{
    // Test case 1
    int arr1[] = {5, 12, 3, 8};
    int n1 = 4;

    cout << "Test 1 - Array: ";
    for (int i = 0; i < n1; i++) 
    {
        cout << arr1[i] << " ";
    }
    cout << endl;

    int sum1 = ArraySum(n1, arr1);
    double avg1 = (double)sum1 / n1;
    cout << "Sum: " << sum1 << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << avg1 << endl;
    cout << endl;

    // Test case 2
    int arr2[] = {10, 20, 30};
    int n2 = 3;

    cout << "Test 2 - Array: ";
    for (int i = 0; i < n2; i++) 
    {
        cout << arr2[i] << " ";
    }
    cout << endl;

    int sum2 = ArraySum(n2, arr2);
    double avg2 = (double)sum2 / n2;
    cout << "Sum: " << sum2 << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << avg2 << endl;
    cout << endl;

    // Test case 3
    int arr3[] = {7};
    int n3 = 1;

    cout << "Test 3 - Array: ";
    for (int i = 0; i < n3; i++) 
    {
        cout << arr3[i] << " ";
    }
    cout << endl;

    int sum3 = ArraySum(n3, arr3);
    double avg3 = (double)sum3 / n3;
    cout << "Sum: " << sum3 << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << avg3 << endl;

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o array_sum array_sum.cpp

To run the program:

./array_sum

Output:

Test 1 - Array: 5 12 3 8 
Sum: 28
Average: 7.00

Test 2 - Array: 10 20 30 
Sum: 60
Average: 20.00

Test 3 - Array: 7 
Sum: 7
Average: 7.00

Modified program with combined function:

#include <iostream>
#include <iomanip>
using namespace std;

// Function to compute both sum and average
void ArraySumAndAverage(int n, int *a, int& sum, double& avg) 
{
    sum = 0;

    for (int i = 0; i < n; i++) 
    {
        sum = sum + a[i];
    }

    avg = (double)sum / n;
}

int main() 
{
    int arr[] = {5, 12, 3, 8};
    int n = 4;
    int sum;
    double average;

    cout << "Array: ";
    for (int i = 0; i < n; i++) 
    {
        cout << arr[i] << " ";
    }
    cout << endl;

    ArraySumAndAverage(n, arr, sum, average);

    cout << "Sum: " << sum << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << average << endl;

    return 0;
}

Output:

Array: 5 12 3 8 
Sum: 28
Average: 7.00

Note: The setprecision(2) manipulator is used to format the average with 2 decimal places. The conversion (double) ensures the floating-point division: without it, sum / n would be the integer division. Since the average divides by \(n\), the functions require \(n \ge 1\).

Exercise 6.4: Introduction to {\tt C++} (E4)

Given an array \(a\) of \(n\) integer numbers and a value \(m\), we consider the problem of determining whether the value \(m\) is present or not in the array.

Questions

Write a complete C++ program that implements the linear search algorithm for an array \(a\) of \(n\) integer numbers. The program should:
- Define a function int LinearSearch(int n, int *a, int m) that searches for value m in array a
- Return the index of the element if found, or -1 if not found
- In the main() function, create a test array and search for different values
- Print the results for each search
Execute the program with the following instance:
- Array: \(a = [9, 12, 3, 16, 19, 11, 12]\) with \(n=7\)
- Search for: \(m=16\) (element present)
- Search for: \(m=25\) (element not present)
- Search for: \(m=9\) (first element)
- Search for: \(m=12\) (duplicate element)
Show the complete compilation commands and output for all test cases.
Modify the program to create a function int CountOccurrences(int n, int *a, int m) that counts how many times value m appears in the array. Execute this modified version with value \(m=12\) in the test array and show the output.

Solution

Complete C++ program:

#include <iostream>
using namespace std;

// Function to perform linear search
int LinearSearch(int n, int *a, int m) 
{
    for (int i = 0; i < n; i++) 
    {
        if (a[i] == m) 
        {
            return i;  // Found, return index
        }
    }

    return -1;  // Not found
}

int main() 
{
    int arr[] = {9, 12, 3, 16, 19, 11, 12};
    int n = 7;

    cout << "Array: ";
    for (int i = 0; i < n; i++) 
    {
        cout << arr[i] << " ";
    }
    cout << endl << endl;

    // Test 1: Search for 16 (present)
    int m1 = 16;
    int result1 = LinearSearch(n, arr, m1);
    cout << "Search for " << m1 << ": ";
    if (result1 != -1) 
    {
        cout << "Found at index " << result1 << endl;
    } 
    else 
    {
        cout << "Not found" << endl;
    }

    // Test 2: Search for 25 (not present)
    int m2 = 25;
    int result2 = LinearSearch(n, arr, m2);
    cout << "Search for " << m2 << ": ";
    if (result2 != -1) 
    {
        cout << "Found at index " << result2 << endl;
    } 
    else 
    {
        cout << "Not found" << endl;
    }

    // Test 3: Search for 9 (first element)
    int m3 = 9;
    int result3 = LinearSearch(n, arr, m3);
    cout << "Search for " << m3 << ": ";
    if (result3 != -1) 
    {
        cout << "Found at index " << result3 << endl;
    } 
    else 
    {
        cout << "Not found" << endl;
    }

    // Test 4: Search for 12 (duplicate)
    int m4 = 12;
    int result4 = LinearSearch(n, arr, m4);
    cout << "Search for " << m4 << ": ";
    if (result4 != -1) 
    {
        cout << "Found at index " << result4 
             << " (first occurrence)" << endl;
    } 
    else 
    {
        cout << "Not found" << endl;
    }

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o linear_search linear_search.cpp

To run the program:

./linear_search

Output:

Array: 9 12 3 16 19 11 12 

Search for 16: Found at index 3
Search for 25: Not found
Search for 9: Found at index 0
Search for 12: Found at index 1 (first occurrence)

Modified program to count occurrences:

#include <iostream>
using namespace std;

// Function to count occurrences of value m
int CountOccurrences(int n, int *a, int m) 
{
    int count = 0;

    for (int i = 0; i < n; i++) 
    {
        if (a[i] == m) 
        {
            count++;
        }
    }

    return count;
}

int main() 
{
    int arr[] = {9, 12, 3, 16, 19, 11, 12};
    int n = 7;

    cout << "Array: ";
    for (int i = 0; i < n; i++) 
    {
        cout << arr[i] << " ";
    }
    cout << endl;

    int x = 12;
    int count = CountOccurrences(n, arr, x);

    cout << "Value " << x << " appears " << count 
         << " times in the array" << endl;

    return 0;
}

Output:

Array: 9 12 3 16 19 11 12 
Value 12 appears 2 times in the array

Note: The LinearSearch function returns the index of the first occurrence when duplicates exist. The CountOccurrences function scans the entire array to count all occurrences.

Exercise 6.5: Introduction to {\tt C++} (E5)

Given an array \(a\) of \(n \ge 1\) integer numbers, we consider the problem of computing in a single pass the maximum, minimum, sum and average of all elements in the array.

Questions

Write a complete C++ program that computes all statistics for an array \(a\) of \(n\) integer numbers in a single pass. The program should:
- Define a function void ArrayStatistics(int n, int *a, int& max, int& min, int& sum, double& avg) that computes all four statistics
- In the main() function, create test arrays and call the function
- Print all statistics for each test case
Execute the program on the following instances:
- Instance 1: \(a = [9, 12, 3, 16, 19, 11, 12]\) with \(n=7\)
- Instance 2: \(a = [5, 2, 8, 1]\) with \(n=4\)
Show the complete compilation commands and output for all test cases.
Discuss the advantages of computing all statistics in a single pass versus computing them separately with multiple functions.

Solution

Complete C++ program:

#include <iostream>
#include <iomanip>
using namespace std;

// Function to compute all statistics in a single pass
void ArrayStatistics(int n, int *a, int& max, int& min, 
                     int& sum, double& avg) 
{
    // Initialize with first element
    max = a[0];
    min = a[0];
    sum = a[0];

    // Process remaining elements
    for (int i = 1; i < n; i++) 
    {
        if (a[i] > max) 
        {
            max = a[i];
        }
        if (a[i] < min) 
        {
            min = a[i];
        }
        sum = sum + a[i];
    }

    // Compute average
    avg = (double)sum / n;
}

int main() 
{
    // Test case 1
    int arr1[] = {9, 12, 3, 16, 19, 11, 12};
    int n1 = 7;
    int max1, min1, sum1;
    double avg1;

    cout << "Test 1 - Array: ";
    for (int i = 0; i < n1; i++) 
    {
        cout << arr1[i] << " ";
    }
    cout << endl;

    ArrayStatistics(n1, arr1, max1, min1, sum1, avg1);

    cout << "Maximum: " << max1 << endl;
    cout << "Minimum: " << min1 << endl;
    cout << "Sum: " << sum1 << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << avg1 << endl;
    cout << endl;

    // Test case 2
    int arr2[] = {5, 2, 8, 1};
    int n2 = 4;
    int max2, min2, sum2;
    double avg2;

    cout << "Test 2 - Array: ";
    for (int i = 0; i < n2; i++) 
    {
        cout << arr2[i] << " ";
    }
    cout << endl;

    ArrayStatistics(n2, arr2, max2, min2, sum2, avg2);

    cout << "Maximum: " << max2 << endl;
    cout << "Minimum: " << min2 << endl;
    cout << "Sum: " << sum2 << endl;
    cout << "Average: " << fixed << setprecision(2) 
         << avg2 << endl;

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o array_stats array_stats.cpp

To run the program:

./array_stats

Output:

Test 1 - Array: 9 12 3 16 19 11 12 
Maximum: 19
Minimum: 3
Sum: 82
Average: 11.71

Test 2 - Array: 5 2 8 1 
Maximum: 8
Minimum: 1
Sum: 16
Average: 4.00

Advantages of single-pass approach:

Advantages:
- Fewer array traversals: The array is traversed only once instead of once for each statistic
- More compact code: Single function and single loop instead of multiple function calls
- Better for streaming data: Can process data that arrives sequentially or doesn't fit entirely in memory
- Reduced overhead: Only one loop initialization and termination instead of multiple loops
Trade-offs:
- Less modular: Combines multiple operations in one function, which may be harder to maintain
- Less flexible: If only one statistic is needed, the function still computes all of them
- Code reusability: Cannot easily reuse individual statistic computations in other contexts
Recommendation: The single-pass approach is preferable when:
- All statistics are needed together
- Working with large arrays where minimizing array accesses is important
- Processing streaming data or data that cannot fit in memory
Separate functions are better when:
- Only specific statistics are needed
- Code clarity and modularity are higher priorities
- Individual functions may be reused in different contexts

Exercise 6.6: Introduction to {\tt C++} (E6)

Given a two-dimensional array \(a\) of \(m \times n\) integer numbers, we consider the problem of computing the sum of all its elements.

Questions

Write a complete C++ program that computes the sum of all elements of a two-dimensional array \(a\) of \(m \times n\) integers. The program should:
- Define a function int SumMatrix(int m, int n, int **matrix) that computes the sum
- In the main() function, allocate test matrices using dynamic memory, initialize them, call the function, and print the results
- Free all allocated memory after use
Execute the program on the following instances:
- Instance 1: a \(3 \times 4\) matrix with sequential values \(a[i][j] = (i-1)\cdot n + j\) for \(i \in \{1, 2, 3\}\) and \(j \in \{1, 2, 3, 4\}\) (rows and columns numbered from \(1\), as in pseudo-code), i.e.,
  
  \[ \begin{bmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\end{bmatrix} \]
- Instance 2: a \(2 \times 3\) matrix with values \(\begin{bmatrix} 3 & 1 & 4 \\ 1 & 5 & 9 \end{bmatrix}\)
Show the complete compilation commands and output for both test cases.
Modify the program to add a function void RowSums(int m, int n, int **matrix, int *rowSums) that computes the sum of each row and stores the results in the array rowSums. Execute this modified version on Instance 1 and show the output.

Solution

Complete C++ program:

#include <iostream>
using namespace std;

// Function to compute the sum of all elements in a matrix
int SumMatrix(int m, int n, int **matrix) 
{
    int sum = 0;
    for (int i = 0; i < m; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            sum += matrix[i][j];
        }
    }
    return sum;
}

int main() 
{
    // Test case 1: 3x4 matrix with sequential values
    int m1 = 3, n1 = 4;

    int** mat1 = new int*[m1];
    for (int i = 0; i < m1; i++) 
    {
        mat1[i] = new int[n1];
    }

    int val = 1;
    for (int i = 0; i < m1; i++) 
    {
        for (int j = 0; j < n1; j++) 
        {
            mat1[i][j] = val++;
        }
    }

    cout << "Test 1 - Matrix " << m1 << "x" << n1 << ":" << endl;
    for (int i = 0; i < m1; i++) 
    {
        for (int j = 0; j < n1; j++) 
        {
            cout << mat1[i][j] << " ";
        }
        cout << endl;
    }
    cout << "Sum: " << SumMatrix(m1, n1, mat1) << endl << endl;

    for (int i = 0; i < m1; i++) 
    {
        delete[] mat1[i];
    }
    delete[] mat1;

    // Test case 2: 2x3 matrix
    int m2 = 2, n2 = 3;
    int data[2][3] = {{3, 1, 4}, {1, 5, 9}};

    int** mat2 = new int*[m2];
    for (int i = 0; i < m2; i++) 
    {
        mat2[i] = new int[n2];
        for (int j = 0; j < n2; j++) 
        {
            mat2[i][j] = data[i][j];
        }
    }

    cout << "Test 2 - Matrix " << m2 << "x" << n2 << ":" << endl;
    for (int i = 0; i < m2; i++) 
    {
        for (int j = 0; j < n2; j++) 
        {
            cout << mat2[i][j] << " ";
        }
        cout << endl;
    }
    cout << "Sum: " << SumMatrix(m2, n2, mat2) << endl;

    for (int i = 0; i < m2; i++) 
    {
        delete[] mat2[i];
    }
    delete[] mat2;

    return 0;
}

Compilation and execution:

To compile the program:

g++ -Wall -std=c++17 -o sum_matrix sum_matrix.cpp

To run the program:

./sum_matrix

Output:

Test 1 - Matrix 3x4:
1 2 3 4 
5 6 7 8 
9 10 11 12 
Sum: 78

Test 2 - Matrix 2x3:
3 1 4 
1 5 9 
Sum: 23

Row sums function:

#include <iostream>
using namespace std;

// Function to compute the sum of all elements in a matrix
int SumMatrix(int m, int n, int **matrix) 
{
    int sum = 0;
    for (int i = 0; i < m; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            sum += matrix[i][j];
        }
    }
    return sum;
}

// Function to compute the sum of each row
void RowSums(int m, int n, int **matrix, int *rowSums) 
{
    for (int i = 0; i < m; i++) 
    {
        rowSums[i] = 0;
        for (int j = 0; j < n; j++) 
        {
            rowSums[i] += matrix[i][j];
        }
    }
}

int main() 
{
    // Test case 1: 3x4 matrix with sequential values
    int m1 = 3, n1 = 4;

    int** mat1 = new int*[m1];
    for (int i = 0; i < m1; i++) 
    {
        mat1[i] = new int[n1];
    }

    int val = 1;
    for (int i = 0; i < m1; i++) 
    {
        for (int j = 0; j < n1; j++) 
        {
            mat1[i][j] = val++;
        }
    }

    cout << "Matrix " << m1 << "x" << n1 << ":" << endl;
    for (int i = 0; i < m1; i++) 
    {
        for (int j = 0; j < n1; j++) 
        {
            cout << mat1[i][j] << " ";
        }
        cout << endl;
    }
    cout << "Total sum: " << SumMatrix(m1, n1, mat1) << endl << endl;

    int* rowSums = new int[m1];
    RowSums(m1, n1, mat1, rowSums);

    cout << "Row sums:" << endl;
    for (int i = 0; i < m1; i++) 
    {
        cout << "Row " << (i + 1) << ": " << rowSums[i] << endl;
    }

    delete[] rowSums;
    for (int i = 0; i < m1; i++) 
    {
        delete[] mat1[i];
    }
    delete[] mat1;

    return 0;
}

Output:

Matrix 3x4:
1 2 3 4 
5 6 7 8 
9 10 11 12 
Total sum: 78

Row sums:
Row 1: 10
Row 2: 26
Row 3: 42