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AI Coding & CSE Fundamentals

AI Coding & CSE Fundamentals

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πŸš€ Explore AI, Coding & Computer Science Fundamentals. πŸ’» Programming β€’ DSA β€’ DBMS β€’ OS β€’ Networks πŸ€– AI & Emerging Technologies πŸ› οΈ Projects β€’ Tutorials β€’ Learning Resources πŸ“š Your freedom to gain knowledge for free.

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πŸ“ˆ Analytical overview of Telegram channel AI Coding & CSE Fundamentals

Channel AI Coding & CSE Fundamentals (@aicoding_csefundamentals) in the English language segment is an active participant. Currently, the community unites 12 324 subscribers, ranking 10 232 in the Technologies & Applications category and 31 271 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 12 324 subscribers.

According to the latest data from 12 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -238 over the last 30 days and by -8 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 10.45%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 288 views. Within the first day, a publication typically gains 0 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 11.
  • Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œπŸš€ Explore AI, Coding & Computer Science Fundamentals. πŸ’» Programming β€’ DSA β€’ DBMS β€’ OS β€’ Networks πŸ€– AI & Emerging Technologies πŸ› οΈ Projects β€’ Tutorials β€’ Learning Resources πŸ“š Your freedom to gain knowledge for free.”

Thanks to the high frequency of updates (latest data received on 13 November, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.

12 324
Subscribers
-824 hours
-827 days
-23830 days
Posts Archive
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸ’» Area of Circle Calculator
#include <stdio.h>

int main() {
    float radius, area;
    const float PI = 3.14159;

    printf("Enter the radius of the circle: ");
    scanf("%f", &radius);

    area = PI * radius * radius;

    printf("The area of the circle is: %fn", area);

    return 0;
}
πŸ“€ Output:
Input: 5
Output: Enter the radius of the circle: The area of the circle is: 78.539757

Input: 2.5
Output: Enter the radius of the circle: The area of the circle is: 19.634938

πŸ’» Compound Interest Calculator
#include <stdio.h>
#include <math.h>

int main() {
  float principal, rate, time, compound_interest, amount;
  int compounding_frequency;

  printf("Enter principal amount: ");
  scanf("%f", &principal);

  printf("Enter annual interest rate (as a decimal, e.g., 0.05 for 5%%): ");
  scanf("%f", &rate);

  printf("Enter time in years: ");
  scanf("%f", &time);

  printf("Enter compounding frequency (e.g., 1 for annually, 12 for monthly): ");
  scanf("%d", &compounding_frequency);

  amount = principal * pow(1 + (rate / compounding_frequency), compounding_frequency * time);
  compound_interest = amount - principal;

  printf("Compound Interest: %.2fn", compound_interest);

  return 0;
}
πŸ“€ Output:
Enter principal amount: 1000
Input: 1000
Enter annual interest rate (as a decimal, e.g., 0.05 for 5%): 0.05
Input: 0.05
Enter time in years: 5
Input: 5
Enter compounding frequency (e.g., 1 for annually, 12 for monthly): 1
Input: 1
Compound Interest: 276.28

πŸ’» Simple Interest Calculator
#include <stdio.h>

int main() {
  float principal, rate, time, interest;

  printf("Enter principal amount: ");
  scanf("%f", &principal);

  printf("Enter rate of interest: ");
  scanf("%f", &rate);

  printf("Enter time in years: ");
  scanf("%f", &time);

  interest = (principal * rate * time) / 100;

  printf("Simple Interest = %fn", interest);

  return 0;
}
πŸ“€ Output:
Input: 1000
Input: 5
Input: 2
Output: Simple Interest = 100.000000

πŸ’» Variable Swapping (Three Variables)
#include <stdio.h>

int main() {
  int a, b, c, temp;

  printf("Enter the values of a, b, and c: ");
  scanf("%d %d %d", &a, &b, &c);

  printf("Before swapping: a = %d, b = %d, c = %dn", a, b, c);

  temp = a;
  a = b;
  b = c;
  c = temp;

  printf("After swapping: a = %d, b = %d, c = %dn", a, b, c);

  return 0;
}
πŸ“€ Output:
Input: 10 20 30
Output: Enter the values of a, b, and c: Before swapping: a = 10, b = 20, c = 30
After swapping: a = 20, b = 30, c = 10

πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸ’» Variable Swapping (Two Variables)
#include <stdio.h>

int main() {
  int a, b, temp;

  printf("Enter the value of a: ");
  scanf("%d", &a);

  printf("Enter the value of b: ");
  scanf("%d", &b);

  printf("Before swapping: a = %d, b = %dn", a, b);

  temp = a;
  a = b;
  b = temp;

  printf("After swapping: a = %d, b = %dn", a, b);

  return 0;
}
πŸ“€ Output:
Input: 5
Input: 10
Output: Enter the value of a: Enter the value of b: Before swapping: a = 5, b = 10
After swapping: a = 10, b = 5

πŸ’» Basic Calculator (Add, Subtract, Multiply, Divide)
#include <stdio.h>

int main() {
    float num1, num2, result;
    char operator;

    printf("Enter first number: ");
    scanf("%f", &num1);

    printf("Enter an operator (+, -, *, /): ");
    scanf(" %c", &operator);

    printf("Enter second number: ");
    scanf("%f", &num2);

    switch (operator) {
        case '+':
            result = num1 + num2;
            printf("%.2f + %.2f = %.2fn", num1, num2, result);
            break;
        case '-':
            result = num1 - num2;
            printf("%.2f - %.2f = %.2fn", num1, num2, result);
            break;
        case '*':
            result = num1 * num2;
            printf("%.2f * %.2f = %.2fn", num1, num2, result);
            break;
        case '/':
            if (num2 == 0) {
                printf("Error! Division by zero.n");
            } else {
                result = num1 / num2;
                printf("%.2f / %.2f = %.2fn", num1, num2, result);
            }
            break;
        default:
            printf("Error! Invalid operator.n");
    }

    return 0;
}
πŸ“€ Output:
Input: 10
Input: +
Input: 5
Output: 10.00 + 5.00 = 15.00

Input: 20
Input: -
Input: 7
Output: 20.00 - 7.00 = 13.00

Input: 4
Input: *
Input: 6
Output: 4.00 * 6.00 = 24.00

Input: 15
Input: /
Input: 3
Output: 15.00 / 3.00 = 5.00

Input: 8
Input: /
Input: 0
Output: Error! Division by zero.

Input: 9
Input: %
Input: 2
Output: Error! Invalid operator.

πŸ’» Hello World Program
#include <stdio.h>

int main() {
    printf("Hello, World!n");
    return 0;
}
πŸ“€ Output:
Hello, World!

πŸ”§ C Basics & Syntax

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

Print days of the week using switch statement
#include <stdio.h>

int main() {
    int dayNumber;

    printf("Enter a number (1-7): ");
    scanf("%d", &dayNumber);

    switch (dayNumber) {
        case 1:
            printf("Sundayn");
            break;
        case 2:
            printf("Mondayn");
            break;
        case 3:
            printf("Tuesdayn");
            break;
        case 4:
            printf("Wednesdayn");
            break;
        case 5:
            printf("Thursdayn");
            break;
        case 6:
            printf("Fridayn");
            break;
        case 7:
            printf("Saturdayn");
            break;
        default:
            printf("Invalid inputn");
    }

    return 0;
}

πŸ’‘ Approach Step 1: Get integer input from the user representing the day of the week (1 for Sunday, 2 for Monday, ..., 7 for Saturday). Step 2: Create a switch statement. The switch expression will be the integer input from the user. Step 3: Inside the switch statement, create case labels for each day of the week (1 to 7). Step 4: Within each case label, use printf to print the corresponding day of the week (e.g., case 1: printf("Sunday"); break;). Include a break statement after each printf to exit the switch statement after a match. Step 5: Add a default case to handle invalid input (numbers outside the range 1-7). Print an error message like "Invalid input". ───────────────────────────── Have you Understood? Drop a reaction: ❀️ Understood | πŸ‘Ž Not Understood

πŸ“ Print days of the week using switch statement Write a C program that takes an integer input representing a day of the week (1 for Sunday, 2 for Monday, etc.). Using a switch statement, print the corresponding day of the week's name based on the input number.

Implement a simple calculator using `switch` statement for operations (+, -, , /) #include <stdio.h> int main() { float num1, num2, result; char operator; printf("Enter first number: "); scanf("%f", &num1); printf("Enter operator (+, -, , /): "); scanf(" %c", &operator); printf("Enter second number: "); scanf("%f", &num2); switch (operator) { case '+': result = num1 + num2; break; case '-': result = num1 - num2; break; case '': result = num1 num2; break; case '/': if (num2 == 0) { printf("Error: Division by zero is not allowed.\n"); return 1; } result = num1 / num2; break; default: printf("Error: Invalid operator.\n"); return 1; } printf("Result: %.2f\n", result); return 0; }

πŸ’‘ Approach Step 1: Include Header and Declare Variables: Include the standard input/output library (`stdio.h`). Declare variables to store the two numbers (e.g., `num1`, `num2` as `float` for decimal support), the operator (e.g., `operator` as `char`), and the result (e.g., `result` as `float`). Step 2: Get User Input: Prompt the user to enter the two numbers and the operator (+, -, , /). Use `scanf` to read these values from the console. Step 3: Implement the `switch` Statement: Use a `switch` statement to perform the calculation based on the operator entered by the user. Each `case` will correspond to a specific operator. Step 4: Handle Each Case: Inside each `case`, perform the corresponding arithmetic operation. For example, in the `case '+'`, calculate `result = num1 + num2`. Step 5: Handle Division by Zero: In the `case '/'`, add a check to prevent division by zero. If `num2` is zero, print an error message and possibly exit the `switch` statement or program. Step 6: Provide a `default` Case: Include a `default` case in the `switch` statement to handle invalid operator inputs. Print an error message if the operator is not recognized. Step 7: Print the Result: After the `switch` statement (assuming no errors occurred), print the calculated `result` to the console with appropriate formatting. Step 8: Return 0:* At the end of the `main` function, return 0 to indicate successful execution of the program. ───────────────────────────── Have you Understood\? Drop a reaction: ❀️ Understood | πŸ‘Ž Not Understood