#include<stdio.h>
using namespace std;

// Recursive function to check if a string is a palindrome
bool isPalindrome(string str, int start, int end) {
    // Base case: If the start index is greater than or equal to the end index
    if (start >= end) {
        return true;
    }
    // Check if the first and last characters are the same
    if (str[start] != str[end]) {
        return false;
    }
    // Recursive case: Check the substring excluding the first and last characters
    return isPalindrome(str, start + 1, end - 1);
}

int main() {
    string input;
    cout << "Enter a string: "; cin >> input;

    if (isPalindrome(input, 0, input.length() - 1)) {
        cout << input << " is a palindrome." << endl;
    } else {
        cout << input << " is not a palindrome." << endl;
    }

    return 0;
}

Sample I/O:
Enter a string: noon
noon is a palindrome.
=== Code Execution Successful ===

🔗 Run Online: Check Palindrome

The problem requires character-wise transformation of a string where each alphabet is shifted by one position in the alphabet. This is commonly known as a Caesar cipher with shift 1.

Logic:

Each character is checked individually. If it is 'z', it wraps to 'a'. If it is 'Z', it wraps to 'A'. Otherwise, ASCII value is incremented by 1.

Pseudocode:

FUNCTION shiftString(S)
 RESULT = "
 FOR EACH character c IN S
 IF c == 'z' THEN
 RESULT = RESULT + 'a'
 ELSE IF c == 'Z' THEN
 RESULT = RESULT + 'A'
 ELSE
 RESULT = RESULT + (c + 1)
 END IF
 END FOR
 RETURN RESULT
END FUNCTION

Conclusion: This algorithm runs in O(n) time complexity where n is the length of the string.

public class Calculator {
    public static int add(int a, int b) {
        return a + b;
    }
    public static void main(String[] args) {
        int result = add(10, 20);
        System.out.println("The sum is: " + result);
    }
}

main() মেথডে

• মেথড যোগফল হিসাব করে ৩০ রিটার্ন করেছে, যেটা result ভ্যারিয়েবলে সংরক্ষণ করা হয়েছে।

• শেষে System.out.println() দিয়ে “The sum is: 30” প্রিন্ট করা হয়েছে।

#include <bits/stdc++.h>
using namespace std;

// function to check if brackets are balanced
bool areBracketsBalanced(string expr)
{
    stack s;
    char x;

    // Traversing the Expression
    for (int i = 0; i < expr.length(); i++) 
    {
        if (expr[i] == '(' || expr[i] == '[' || expr[i] == '{') 
        {
            // Push the element in the stack
            s.push(expr[i]);
            continue;
        }

        // IF current current character is not opening
        // bracket, then it must be closing. So stack
        // cannot be empty at this point.
        if (s.empty())
            return false;

        switch (expr[i]) {
        case ')':
            // Store the top element in a
            x = s.top();
            s.pop();
            if (x == '{' || x == '[')
                return false;
            break;

        case '}':
            // Store the top element in b
            x = s.top();
            s.pop();
            if (x == '(' || x == '[')
                return false;
            break;

        case ']':
            // Store the top element in c
            x = s.top();
            s.pop();
            if (x == '(' || x == '{')
                return false;
            break;
        }
    }

    // Check Empty Stack
    return (s.empty());
}

// Driver code
int main()
{
    string expr;

    // Taking input from the user
    cout << "Enter an expression: "; cin >> expr;

    // Function call
    if (areBracketsBalanced(expr))
        cout << "Balanced\n";
    else
        cout << "Not Balanced\n";
    return 0;
}

Source : GeeksforGeeks

#include <bits/stdc++.h>
using namespace std;

#define ll long long

bool isMountain(vector<ll> a) {
    ll n = a.size();

    if (n < 3)
        return false;

    ll i = 1;

    // Increasing part
    while (i < n && a[i] > a[i - 1]) {
        i++;
    }

    // Peak cannot be first or last
    if (i == 1 || i == n)
        return false;

    // Decreasing part
    while (i < n && a[i] < a[i - 1]) {
        i++;
    }

    return i == n;
}

int main() {
    ll n;
    cin >> n;

    vector<ll> a(n);
    for (ll i = 0; i < n; i++) {
        cin >> a[i];
    }

    if (isMountain(a)) {
        cout << "Mountain" << endl;
    } else {
        cout << "Not Mountain" << endl;
    }

    return 0;
}

Source: Geeksforgeeks

#include <bits/stdc++.h>
using namespace std;

#define N 6

// Define structure for Activity
struct Activity {
    int start, finish;
};

// Function to compare two activities
bool Sort_activity(Activity s1, Activity s2) {
    return (s1.finish < s2.finish); // Sort by finish time
}

// Function to print the maximum number of activities
void print_Max_Activities(Activity arr[], int n) {
    // Sort activities based on finish time
    sort(arr, arr + n, Sort_activity);

    cout << "Following activities are selected: \n";

    int i = 0;
    cout << "(" << arr[i].start << ", " << arr[i].finish << ")\n";

    // Select the next activity that starts after the last selected activity finishes
    for (int j = 1; j < n; j++) {
        if (arr[j].start >= arr[i].finish) {
            cout << "(" << arr[j].start << ", " << arr[j].finish << ")\n";
            i = j;
        }
    }
}

int main() {
    Activity arr[N];

    // Taking input from the user
    for (int i = 0; i < N; i++) {
        cout << "Enter the start and finish time for job " << i + 1 << " : ";
        cin >> arr[i].start >> arr[i].finish;
    }

    // Function call to print the selected activities
    print_Max_Activities(arr, N);

    return 0;
}

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

// Bank Management System using Class & Inheritance in C++
/*
1. Saving Account
2. Current Account
3. Account Creation
4. Deposit
5. Withdraw
6. Balance
*/

class account {
private:
    string name;
    int accno;
    string atype;

public:
    void getAccountDetails() {
        cout << "\nEnter Customer Name : ";
        cin >> name;
        cout << "Enter Account Number : ";
        cin >> accno;
        cout << "Enter Account Type : ";
        cin >> atype;
    }

    void displayDetails() {
        cout << "\n\nCustomer Name : " << name;
        cout << "\nAccount Number : " << accno;
        cout << "\nAccount Type : " << atype;
    }
};

class current_account : public account {
private:
    float balance = 0;

public:
    void c_display() {
        cout << "\nBalance : " << balance;
    }

    void c_deposit() {
        float deposit;
        cout << "\nEnter amount to Deposit : ";
        cin >> deposit;
        balance = balance + deposit;
    }

    void c_withdraw() {
        float withdraw;
        cout << "\n\nBalance : " << balance;
        cout << "\nEnter amount to be withdraw : ";
        cin >> withdraw;

        if (balance >= withdraw && balance > 1000) {
            balance = balance - withdraw;
            cout << "\nBalance Amount After Withdraw: " << balance;
        } else {
            cout << "\nInsufficient Balance";
        }
    }
};

class saving_account : public account {
private:
    float sav_balance = 0;

public:
    void s_display() {
        cout << "\nBalance : " << sav_balance;
    }

    void s_deposit() {
        float deposit, interest;
        cout << "\nEnter amount to Deposit : ";
        cin >> deposit;
        sav_balance = sav_balance + deposit;
        interest = (sav_balance * 2) / 100;
        sav_balance = sav_balance + interest;
    }

    void s_withdraw() {
        float withdraw;
        cout << "\nBalance : " << sav_balance;
        cout << "\nEnter amount to be withdraw : ";
        cin >> withdraw;

        if (sav_balance >= withdraw && sav_balance > 500) {
            sav_balance = sav_balance - withdraw;
            cout << "\nBalance Amount After Withdraw: " << sav_balance;
        } else {
            cout << "\nInsufficient Balance";
        }
    }
};

int main() {
    current_account c1;
    saving_account s1;
    char type;
    int choice;

    cout << "\nEnter S for saving customer and C for current a/c customer : ";
    cin >> type;

    if (type == 's' || type == 'S') {
        s1.getAccountDetails();

        while (1) {
            cout << "\nChoose Your Choice" << endl;
            cout << "1) Deposit" << endl;
            cout << "2) Withdraw" << endl;
            cout << "3) Display Balance" << endl;
            cout << "4) Display with full Details" << endl;
            cout << "5) Exit" << endl;
            cout << "Enter Your choice: ";
            cin >> choice;

            switch (choice) {
                case 1:
                    s1.s_deposit();
                    break;
                case 2:
                    s1.s_withdraw();
                    break;
                case 3:
                    s1.s_display();
                    break;
                case 4:
                    s1.displayDetails();
                    s1.s_display();
                    break;
                case 5:
                    goto end;
                default:
                    cout << "\n\nEntered choice is invalid, TRY AGAIN";
            }
        }
    } 
    else if (type == 'c' || type == 'C') {
        c1.getAccountDetails();

        while (1) {
            cout << "\nChoose Your Choice" << endl;
            cout << "1) Deposit" << endl;
            cout << "2) Withdraw" << endl;
            cout << "3) Display Balance" << endl;
            cout << "4) Display with full Details" << endl;
            cout << "5) Exit" << endl;
            cout << "Enter Your choice: ";
            cin >> choice;

            switch (choice) {
                case 1:
                    c1.c_deposit();
                    break;
                case 2:
                    c1.c_withdraw();
                    break;
                case 3:
                    c1.c_display();
                    break;
                case 4:
                    c1.displayDetails();
                    c1.c_display();
                    break;
                case 5:
                    goto end;
                default:
                    cout << "\n\nEntered choice is invalid, TRY AGAIN";
            }
        }
    } 
    else {
        cout << "\nInvalid Account Selection";
    }

end:
    cout << "\nThank You for Banking with us..";
    return 0;
}

#include <stdio.h>
/* Function to calculate sum of even numbers from 1 to n */
int sumOfEven(int n) {
    int sum = 0;
    for (int i = 2; i <= n; i += 2) {
        sum += i;
    }
    return sum;
}
int main() {
    int n, result;
    printf("Enter an integer: ");
    scanf("%d", &n);
    result = sumOfEven(n);
    printf("Sum of even numbers from 1 to %d = %d\n", n, result);
    return 0;
}

    
Sample I/O:
  Enter an integer: 50
Sum of even numbers from 1 to 50 = 650
=== Code Execution Successful ===

#include <iostream>
using namespace std;

class BankAccount {
private:
    string name;
    int accountNumber;
    double balance;

public:
    void setDetails(string n, int accNo, double bal) {
        name = n;
        accountNumber = accNo;
        balance = bal;
    }

    void deposit(double amount) {
        balance += amount;
        cout << "Deposited: " << amount << endl;
    }

    void withdraw(double amount) {
        if (amount <= balance) {
            balance -= amount;
            cout << "Withdrawn: " << amount << endl;
        } else {
            cout << "Insufficient balance" << endl;
        }
    }
    void display() {
        cout << "\nAccount Details\n";
        cout << "Account Holder: " << name << endl;
        cout << "Account Number: " << accountNumber << endl;
        cout << "Balance: " << balance << endl;
    }
};
int main() {
    BankAccount acc;
    string name;
    int accNo;
    double bal, dep, wit;
    cout << "Enter account holder name: ";
    getline(cin, name);

    cout << "Enter account number: ";
    cin >> accNo;
    cout << "Enter initial balance: ";
    cin >> bal;
    acc.setDetails(name, accNo, bal);
    cout << "Enter amount to deposit: ";
    cin >> dep;
    acc.deposit(dep);
    cout << "Enter amount to withdraw: ";
    cin >> wit;
    acc.withdraw(wit);
    acc.display();
    return 0;
}

    
        Sample I/O:
 Enter account holder name: itjobqns
Enter account number: 02252016
Enter initial balance: 5000
Enter amount to deposit: 300
Deposited: 300
Enter amount to withdraw: 100
Withdrawn: 100
Account Details
Account Holder: itjobqns
Account Number: 2252016
Balance: 5200
=== Code Execution Successful ===

Difference between Exception and Error in Java

• Meaning: Exception represents conditions that a program can handle; Error represents serious problems that a program cannot handle.
• Recovery: Exceptions can be caught and handled using try-catch; Errors are generally not recoverable.
• Cause: Exceptions occur due to logical issues or user input; Errors occur due to system-level failures.
• Examples: Exception – NullPointerException, ArithmeticException; Error – OutOfMemoryError, StackOverflowError.
• Package: Both belong to java.lang package, but Error is not meant to be handled.

Exception Handling

Exception handling is a mechanism used in programming to handle runtime errors so that the normal flow of the program is not interrupted. It helps detect errors and provides a way to recover from them.

Example (Java):

try {
    int a = 10;
    int b = 0;
    int c = a / b;
    System.out.println(c);
} catch (ArithmeticException e) {
    System.out.println("Division by zero is not allowed");
}

Here, the exception is caught and handled instead of crashing the program.

Examples:

• High-Level Languages: C, C++, Java, Python, etc.
• Low-Level Languages: Assembly Language, Machine Code.

#include <iostream>
using namespace std;
void matrix_multiply(int A[10][10], int B[10][10], int C[10][10], int m, int n, int p, int q) {
if (n != p) {
cout << "Matrices A and B cannot be multiplied due to incompatible dimensions.\n";
return;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < q; j++) {
C[i][j] = 0;
for (int k = 0; k < n; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
}
float calculate_average(int matrix[10][10], int m, int n) {
int sum = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sum += matrix[i][j];
}
}
return static_cast(sum) / (m * n);
}
int find_max(int matrix[10][10], int m, int n) {
int max = matrix[0][0];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) { if (matrix[i][j] > max) {
max = matrix[i][j];
}
}
}
return max;
}
int main() {
int A[10][10], B[10][10], C[10][10];
int m, n, p, q;
cout << "Enter the number of rows and columns for matrix A (m x n): "; cin >> m >> n;
cout << "Enter the number of rows and columns for matrix B (p x q): "; cin >> p >> q;
if (n != p) {
cout << "Matrices A and B cannot be multiplied due to incompatible dimensions.\n";
return 0;
}

// Input matrices A and B
cout << "Enter elements of matrix A:\n";
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) { cin >> A[i][j];
}
}
cout << "Enter elements of matrix B:\n";
for (int i = 0; i < p; i++) {
for (int j = 0; j < q; j++) { cin >> B[i][j];
}
}
// A) Multiply matrices A and B
matrix_multiply(A, B, C, m, n, p, q);
cout << "Matrix C (A x B):\n";
for (int i = 0; i < m; i++) {
for (int j = 0; j < q; j++) {
cout << C[i][j] << " ";
}
cout << endl;
}

// B) Calculate the average of matrices A and B
float average_A = calculate_average(A, m, n);
float average_B = calculate_average(B, p, q);
cout << "Average of Matrix A: " << average_A << endl;
cout << "Average of Matrix B: " << average_B << endl;
// C) Find the maximum value in matrix C
int max_C = find_max(C, m, q);
cout << "Maximum value in Matrix C: " << max_C << endl;
return 0;
}

   Sample I/O:
        Enter the number of rows and columns for matrix A (m x n): 3 3
Enter the number of rows and columns for matrix B (p x q): 3 3
Enter elements of matrix A:
1 5 7 6 3 2 69 7 12
Enter elements of matrix B:
45 63 2 9 8 7 12 10 3
Matrix C (A x B):
174 173 58 
321 422 39 
3312 4523 223 
Average of Matrix A: 12.4444
Average of Matrix B: 17.6667
Maximum value in Matrix C: 4523
=== Code Execution Successful ===
        

#include <iostream>
using namespace std;

int singleDigit(int n) {
    int sum = n;

    // Repeatedly calculate sum of digits until it becomes a single digit
    while (sum >= 10) {
        int tempSum = 0;

        // Calculate sum of digits of the number
        while (sum > 0) {
            tempSum += sum % 10;
            sum /= 10;
        }

        sum = tempSum;  // Update sum to the sum of digits
    }

    return sum;
}

int main() {
    int n;

    // Taking user input
    cout << "Enter a number: ";
    cin >> n;

    // Calling the function and printing the result
    cout << "The single digit sum is: " << singleDigit(n) << endl;

    return 0;
}

Input: n = 1234
Output: 1
Explanation:
Step 1: 1 + 2 + 3 + 4 = 10
Step 2: 1 + 0 = 1

Input: n = 5674
Output: 4
Explanation:
Step 1: 5 + 6 + 7 + 4 = 22
Step 2: 2 + 2 = 4

543210

Output:

8   4   2

• sizeof(6.5): 6.5 is a double in C, and the size of a double is typically 8 bytes. Thus, it will print 8.
• sizeof(90000): 90000 is an integer, and the size of an integer in C is typically 4 bytes. Thus, it will print 4.
• sizeof("A"): "A" is a string literal, and a string in C is stored as an array of characters, including the null terminator. Since "A" consists of 2 characters (A and '\0'), it will print 2.

Algorithm to Find the Smallest Element in an Array

START

Step 1 → Initialize an array A with given values.
Step 2 → Create a variable min_value and set it to a large number (e.g., infinity) or the first element of the array.
Step 3 → Loop through each element A[i] in the array starting from the first element.
Step 4 → For each element, check if A[i] is smaller than min_value.
Step 5 → If A[i] < min_value, update min_value to A[i].
Step 6 → Repeat Steps 4 and 5 until all elements have been checked.
Step 7 → Display min_value as the smallest element of the array.

STOP

#include <stdio.h>

int findSmallest(int arr[], int size) {
    
    // 1. Assume the first number is the smallest
    int smallest = arr[0];
    
    // 2. Look at the rest of the numbers one by one
    for (int i = 1; i < size; ++i) {
        
        // 3. If we find a number that is smaller, save it
        if (arr[i] < smallest) {
            smallest = arr[i];
        }
    }
    
    // 4. Send back the smallest number we found
    return smallest;
}

#include <iostream>
using namespace std;
int main() {
    int i, num, n, count;
    cout << "Enter the range: "; 
    cin >> n;
    cout << "The prime numbers in between the range 1 to " << n << " are: ";

    for (num = 1; num <= n; num++) {
        count = 0;
        for (i = 2; i <= num / 2; i++) {
            if (num % i == 0) {
                count++;
                break;
            }
        }
        if (count == 0 && num != 1)
            cout << num << " ";
    }

    return 0;
}

Sample I/O:
Enter the range: 50
The prime numbers in between the range 1 to 50 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 

=== Code Execution Successful ===

🔗 Run Online: Prime number between range

#include <iostream>
using namespace std;
int main() {
    int i, j, n, p, q;
    cout << "Input number of rows: "; 
    cin >> n;
    for (i = 1; i <= n; i++) {
        if (i % 2 == 0) {
            p = 1;
            q = 0;
        } else {
            p = 0;
            q = 1;
        }
        for (j = 1; j <= i; j++) {
            if (j % 2 == 0)
                cout << p;
            else
                cout << q;
        }
        cout << endl;
    }
    return 0;
}

Sample I/O:
Input number of rows: 6
1
01
101
0101
10101
010101
=== Code Execution Successful ===

#include <iostream>
using namespace std;
int main() {
    int n, i, sum = 0, term = 1, nextTerm;
    cout << "Enter the value of n: "; 
    cin >> n;
    cout << "The series is: ";
    for (i = 1; i <= n; i++) {
        cout << term << " ";
        sum += term;
        nextTerm = term + i;
        term = nextTerm;
    }

    cout << "\nThe sum of the series is: " << sum << endl;
    return 0;
}

Sample I/O:
Enter the value of n: 10
The series is: 1 2 4 7 11 16 22 29 37 46 
The sum of the series is: 175

=== Code Execution Successful ===

Step-by-Step Execution

1. Initial call: fun(5)

   • x = 5
   • Print 5
   • x-- (post-decrement) → Pass x = 4 to fun(--x)

2. Recursive call: fun(3)

   • x = 3
   • Print 3
   • x-- (post-decrement) → Pass x = 2 to fun(--x)

3. Recursive call: fun(1)

   • x = 1
   • Print 1
   • x-- (post-decrement) → Pass x = 0 to fun(--x)

4. Recursive call: fun(-1)

   • x = -1
   • The if(x<0) return; triggers, stopping recursion.
   • Nothing is printed here.

5. Returning from fun(1):

   • x = -1 at this point (because x was decremented before recursion)
   • Print -1

6. Returning from fun(3):

   • x = 1
   • Print 1

7. Returning from fun(5):

   • x = 3
   • Print 3

#include <stdio.h>
int main () {
int n1, n2, max;
printf ("Enter two positive integers: ");
scanf ("%d %d", &n1, &n2);
// maximum number between n1 and n2 is stored in max

max = (n1> n2) ? n1: n2;
while (1) {
// Check if max is divisible by both n1 and n2
if ((max% n1 == 0)&& (max % n2 ==0)) {
 // If true, max is the LCM of n1 and n2
printf("The LCM of %d and %d is %d.", n1,n2,max);
break;
}
++max;
}
return 0;
}

    
Sample I/O:
Enter two positive integers: 14 8
The LCM of 14 and 8 is 56.
=== Code Execution Successful ===

#include <bits/stdc++.h>
using namespace std;

int main() {
    string s;
    cout << "Enter String: ";
    cin >> s;
    int n = s.length();
    int start = 0, end = n - 1;

    while (start < end) {
        // XOR-based swapping without using extra storage
        s[start] ^= s[end];
        s[end] ^= s[start];
        s[start] ^= s[end];
        start++;
        end--;
    }

    cout << "Reverse String = " << s << endl;
    return 0;
}

Sample I/O:
Enter String: itjobqns
Reverse String = snqbojti
=== Code Execution Successful ===

🔗 Run Online: Reverse String

#include <stdio.h>
int main() {
    int n, i;
    float x, sum = 1, term = 1;
    printf("Enter value of x: ");
    scanf("%f", &x);
    printf("Enter number of terms: ");
    scanf("%d", &n);
    // Calculate series
    for (i = 1; i <= n; i++) {
        term = term * x / i;   // x^i / i!
        sum = sum + term;
    }
    printf("Value of e^x = %.4f\n", sum);
    return 0;
}

#include <iostream>
using namespace std;

int main() {
    // Loop from 1 to 100
    for (int n = 1; n <= 100; n++) {
        // Inner loop to repeat 'n' exactly 'n' times
        for (int i = 0; i < n; i++) {
            cout << n << " ";
        }
    }
    cout << endl; // Print a newline at the end of the sequence
    return 0;
}

#include <stdio.h>
void printDivisibleBy3(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        if (arr[i] % 3 == 0) {
            printf("%d\n", arr[i]);
        }
    }
}
int main() {
    int size;
    printf("Enter the number of elements: ");
    scanf("%d", &size);

    int numbers[size];  // Variable Length Array (VLA)
    printf("Enter %d numbers:\n", size);
    for (int i = 0; i < size; i++) {
        scanf("%d", &numbers[i]);
    }
    printf("Numbers divisible by 3:\n");
    printDivisibleBy3(numbers, size);
    return 0;
}

Sample I/O:
Enter the number of elements: 6
Enter 6 numbers:
5 6 9 12 8 11
Numbers divisible by 3:
6
9
12
=== Code Execution Successful ===

Code:

#include <stdio.h>

int main() {
    int ledger[5] = {10, 20, 30, 40, 50};

    int *ptr1 = (int *)(&ledger + 1);
    int *ptr2 = (int *)(ledger + 1);

    printf("%d, %d", *(ptr1 - 1), *(ptr2 + 2));
    return 0;
}

Output:
50, 40

Step-by-Step Explanation

1. Array Layout in Memory

2. ptr1 = (int *)(&ledger + 1)

• &ledger is a pointer to the entire array, type int (*)[5].
• &ledger + 1 moves forward by the size of the whole array (5 ints = 20 bytes), landing just after the array.
• ptr1 now points to memory right after ledger[4].

3. ptr2 = (int *)(ledger + 1)

• ledger decays to pointer to first element (int *).
• ledger + 1 moves to ledger[1], which contains 20.
• ptr2 points to ledger[1].

4. *(ptr1 - 1)

• ptr1 is just past the array end.
• ptr1 - 1 moves back to ledger[4].
• Value = 50.

5. *(ptr2 + 2)

• ptr2 points to ledger[1].
• ptr2 + 2 lands on ledger[3].
• Value = 40.

✅ C Program to Compute Row and Column Sum of a Matrix

#include <stdio.h>
int main() {
    int m, n, i, j;

    // Input matrix dimensions
    printf("Enter the number of rows (m): ");
    scanf("%d", &m);
    printf("Enter the number of columns (n): ");
    scanf("%d", &n);

    int matrix[m][n];

    // Input matrix elements
    printf("Enter the elements of the matrix row-wise:\n");
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) {
            scanf("%d", &matrix[i][j]);
        }
    }

    // Calculate and print row sums
    printf("\nMatrix with Row Sums:\n");
    for (i = 0; i < m; i++) {
        int rowSum = 0;
        for (j = 0; j < n; j++) {
            printf("%d ", matrix[i][j]);
            rowSum += matrix[i][j];
        }
        printf("%d\n", rowSum); // Row sum
    }

    // Calculate and print column sums
    printf("\nColumn Sums:\n");
    for (j = 0; j < n; j++) {
        int colSum = 0;
        for (i = 0; i < m; i++) {
            colSum += matrix[i][j];
        }
        printf("%d ", colSum);
    }

    return 0;
}

Sample I/O:Enter the number of rows (m): 3
Enter the number of columns (n): 4
Enter the elements of the matrix row-wise:
1  3  4  2
2  4  5  3
3  2  2  5
Output:
Matrix with Row Sums:
1  3  4  2  10
2  4  5  3  14
3  2  2  5  12

Column Sums:
6  9  11  10

#include <iostream>
using namespace std;
bool isPrime(int num) {
    if (num < 2) return false;
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) return false;
    }
    return true;
}

void findPrimes(int n) {
    cout << "Prime Numbers: ";
    for (int i = 1; i <= n; i++) {
        if (isPrime(i))
            cout << i << " ";
    }
    cout << endl;
}

int main() {
    int n;
    cout << "Enter value of n: ";
    cin >> n;
    findPrimes(n);
    return 0;
}

Sample I/O:Enter value of n: 20
Prime Numbers: 2 3 5 7 11 13 17 19

✅ Identifying Data and Methods

Data Requirements (Attributes):

• x: X-coordinate of the center.
• y: Y-coordinate of the center.
• radius: Radius of the circle.

Method Requirements (Functions):

• Constructor: Initializes the circle with given coordinates and radius.
• translate(dx, dy): Moves the circle to a new position by updating its coordinates.
• getPerimeter(): Computes the perimeter using the formula: \[ P = 2\pi r \]
• getArea(): Computes the area using the formula: \[ A = \pi r^2 \]
• printCircle(): Displays the current position and radius of the circle.

✅ Implementation in C++

#include <iostream>

#include <cmath>



class Circle {

private:

    double x, y, radius;



public:

    // Constructor

    Circle(double x, double y, double radius) {

        this->x = x;

        this->y = y;

        this->radius = (radius > 0) ? radius : 1.0;

    }



    // Translate method

    void translate(double dx, double dy) {

        x += dx;

        y += dy;

    }



    // Compute perimeter

    double getPerimeter() const {

        return 2 * M_PI * radius;

    }



    // Compute area

    double getArea() const {

        return M_PI * radius * radius;

    }



    // Print circle details

    void printCircle() const {

        std::cout << "Circle at (" << x << ", " << y << ") with radius "

                  << radius << std::endl;

    }

};



// Main function

int main() {

    Circle c(0.0, 0.0, 5.0);

    c.printCircle();



    std::cout << "Perimeter: " << c.getPerimeter() << std::endl;

    std::cout << "Area: " << c.getArea() << std::endl;



    c.translate(3.0, 4.0);

    std::cout << "After translation: " << std::endl;

    c.printCircle();



    return 0;

}

✅ Data Flow of the translate Method

Input:

Translation values dx and dy are passed to the translate method.

Process:

• The method adds dx to the current X-coordinate.
• The method adds dy to the current Y-coordinate.

Output:

The updated coordinates (x', y') reflect the new position of the circle.

The output of the program will be 5.

Code Analysis

1. Function F(n):
   • The function calculates the Fibonacci number for a given n using recursion.
   • Base Cases:
     • If n == 0, it returns 0.
     • If n == 1, it returns 1.
   • Recursive Case:
     • For other values of n, it returns F(n-2) + F(n-1).
2. Main Function:
   • The main function calls F(5) and stores the result (though it should be assigned to a variable like result).

Calculation of F(5)

The Fibonacci sequence calculated by the function is:

F(5) = F(4) + F(3)
F(4) = F(3) + F(2)
F(3) = F(2) + F(1)
F(2) = F(1) + F(0)

Breaking it down:

F(2) = 1 (F(1) + F(0) = 1 + 0)

F(3) = 2 (F(2) + F(1) = 1 + 1)

F(4) = 3 (F(3) + F(2) = 2 + 1)

F(5) = 5 (F(4) + F(3) = 3 + 2)

#include <iostream>
using namespace std;
int main() {
    int size1, size2;
    // Taking input size of first array
    cout << "Enter size of first array: "; 
    cin >> size1;
    int arr1[size1];
    cout << "Enter elements of first array:\n";
    for (int i = 0; i < size1; i++) {
        cin >> arr1[i];
    }
    // Taking input size of second array
    cout << "Enter size of second array: ";
    cin >> size2;
    int arr2[size2];
    cout << "Enter elements of second array:\n";
    for (int i = 0; i < size2; i++) {
        cin >> arr2[i];
    }
    // To store common elements
    int commonElements[size1 < size2 ? size1 : size2];
    int count = 0;
    // Check each element in arr1 against arr2
    for (int i = 0; i < size1; i++) {
        bool found = false;
        for (int j = 0; j < size2; j++) {
            if (arr1[i] == arr2[j]) {
                found = true;
                break;
            }
        }
        if (found) {
            // Check for duplicates
            bool alreadyExists = false;
            for (int k = 0; k < count; k++) {
                if (commonElements[k] == arr1[i]) {
                    alreadyExists = true;
                    break;
                }
            }
            if (!alreadyExists) {
                commonElements[count] = arr1[i];
                count++;
            }
        }
    }
    // Print common elements
    cout << "Common elements: ";
    for (int i = 0; i < count; i++) {
        cout << commonElements[i] << " ";
    }
    cout << endl;
    // Print total count
    cout << "Total number of common elements: " << count << endl;
    return 0;
}

Sample I/O:
Enter size of first array: 5
Enter elements of first array: 0 20 30 40 50
Enter size of second array: 6
Enter elements of second array: 10 2 20 30 60 90
Common elements: 20 30 
Total number of common elements: 2

=== Code Execution Successful ===