AI Coding & CSE Fundamentals
🚀 Explore AI, Coding & Computer Science Fundamentals. 💻 Programming • DSA • DBMS • OS • Networks 🤖 AI & Emerging Technologies 🛠️ Projects • Tutorials • Learning Resources 📚 Your freedom to gain knowledge for free.
Show more📈 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 12 318 subscribers, ranking 10 235 in the Technologies & Applications category and 31 257 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 12 318 subscribers.
According to the latest data from 13 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -236 over the last 30 days and by -9 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 10.85%. 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 1 337 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.
📝 Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
“🚀 Explore AI, Coding & Computer Science Fundamentals.
💻 Programming • DSA • DBMS • OS • Networks
🤖 AI & Emerging Technologies
🛠️ Projects • Tutorials • Learning Resources
📚 Your freedom to gain knowledge for free.”
Thanks to the high frequency of updates (latest data received on 14 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.
#include <stdio.h>
int main() {
int choice, a, b, result;
printf("1. Add\n2. Subtract\n3. Multiply\n4. Divide\nEnter your choice: ");
scanf("%d", &choice);
printf("Enter two numbers: ");
scanf("%d %d", &a, &b);
switch (choice) {
case 1:
result = a + b;
printf("Result: %d\n", result);
break;
case 2:
result = a - b;
printf("Result: %d\n", result);
break;
case 3:
result = a * b;
printf("Result: %d\n", result);
break;
case 4:
if (b != 0) {
result = a / b;
printf("Result: %d\n", result);
} else {
printf("Cannot divide by zero!\n");
}
break;
default:
printf("Invalid choice!\n");
}
return 0;
}#include <stdio.h>
int main() {
int num1, num2;
scanf("%d %d", &num1, &num2);
if ((num1 ^ num2) == 0) {
printf("Numbers are equal!");
} else {
printf("Numbers are not equal!");
}
return 0;
}#include <stdio.h>
#include <math.h>
int main() {
float a, b, c, discriminant, root1, root2, realPart, imagPart;
printf("Enter coefficients a, b, and c: ");
scanf("%f %f %f", &a, &b, &c);
discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("Root 1 = %.2f and Root 2 = %.2f", root1, root2);
} else if (discriminant == 0) {
root1 = root2 = -b / (2 * a);
printf("Root 1 = Root 2 = %.2f;", root1);
} else {
realPart = -b / (2 * a);
imagPart = sqrt(-discriminant) / (2 * a);
printf("Root 1 = %.2f+%.2fi and Root 2 = %.2f-%.2fi", realPart, imagPart, realPart, imagPart);
}
return 0;
}#include <stdio.h>
int main() {
int year;
scanf("%d", &year);
if (year % 4 == 0) {
if (year % 100 == 0) {
if (year % 400 == 0) {
printf("Leap year\n");
} else {
printf("Not a leap year\n");
}
} else {
printf("Leap year\n");
}
} else {
printf("Not a leap year\n");
}
return 0;
}#include <stdio.h>
int main() {
int n, i, isPrime = 1;
scanf("%d", &n);
if (n <= 1) {
isPrime = 0;
} else {
if (n == 2) {
isPrime = 1;
} else {
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
isPrime = 0;
break;
}
}
}
}
if (isPrime)
printf("Prime");
else
printf("Not Prime");
return 0;
}#include <stdio.h>
int main() {
int num1, num2, num3, largest;
scanf("%d %d %d", &num1, &num2, &num3);
if (num1 >= num2 && num1 >= num3) {
largest = num1;
} else if (num2 >= num1 && num2 >= num3) {
largest = num2;
} else {
largest = num3;
}
printf("%d", largest);
return 0;
}
int age = 20;
if (age >= 18) {
printf("You are an adult! ✅n");
}
**2. The `else` Statement: Providing an Alternative**
The `else` statement is used in conjunction with `if` to provide an alternative code block to execute when the `if` condition is false.
`if (condition) {
// Code to execute if the condition is true
} else {
// Code to execute if the condition is false
}`
Example:
int age = 15;
if (age >= 18) {
printf("You are an adult! ✅n");
} else {
printf("You are not an adult yet. ⏳n");
}
**3. The `else if` Statement: Checking Multiple Conditions**
The `else if` statement allows you to check multiple conditions in a sequence.
`if (condition1) {
// Code to execute if condition1 is true
} else if (condition2) {
// Code to execute if condition1 is false AND condition2 is true
} else {
// Code to execute if all conditions are false
}`
Example:
int score = 75;
if (score >= 90) {
printf("Grade: A 🥇n");
} else if (score >= 80) {
printf("Grade: B 🥈n");
} else if (score >= 70) {
printf("Grade: C 🥉n");
} else {
printf("Grade: D or F 😥n");
}
**4. The `switch` Statement: Efficient Multi-Way Branching**
The `switch` statement provides a clean way to select one code block to execute from several options based on the value of an expression.
`switch (expression) {
case value1:
// Code to execute if expression == value1
break;
case value2:
// Code to execute if expression == value2
break;
default:
// Code to execute if expression doesn't match any of the cases
}`
- `expression`: An integer or character expression.
- `case`: Each `case` represents a specific value that `expression` might have.
- `break`: The `break` statement is crucial. It exits the `switch` statement after a match is found. Without `break`, the code will "fall through" to the next `case`. ⚠️
- `default`: The `default` case is optional and is executed if none of the other `case` values match the `expression`.
Example:
int day = 3;
switch (day) {
case 1:
printf("Mondayn");
break;
case 2:
printf("Tuesdayn");
break;
case 3:
printf("Wednesdayn");
break;
default:
printf("Invalid dayn");
}
💡 **Tips for Using Control Flow:**
- Keep your conditions clear and easy to understand.
- Use indentation to make your code readable. ✅
- Always include a `default` case in your `switch` statement to handle unexpected values.
- Be careful about "fall-through" in `switch` statements. Use `break` unless you specifically want this behavior. ⚠️
- When dealing with complex conditions, consider using logical operators (`&&` for AND, `||` for OR, `!` for NOT).
Control flow statements are essential for writing programs that can respond to different situations. Practice using `if`, `else`, and `switch` to master decision-making in your C programs! 💪#include <stdio.h>
int main() {
int x, y, max, min;
scanf("%d %d", &x, &y);
int diff = x - y;
int sign_bit = diff >> 31 & 1;
max = x - sign_bit * diff;
min = y + sign_bit * diff;
printf("Max: %d\n", max);
printf("Min: %d\n", min);
return 0;
}#include <stdio.h>
int main() {
int a = 10, b = 5;
a = a ^ b;
b = a ^ b;
a = a ^ b;
printf("a = %d, b = %d\n", a, b);
return 0;
}#include <stdio.h>
int main() {
unsigned int num = 10;
int bit_position = 1;
printf("Original number: %u\n", num);
unsigned int set_bit = num | (1 << bit_position);
printf("Number with bit set: %u\n", set_bit);
unsigned int clear_bit = num & ~(1 << bit_position);
printf("Number with bit cleared: %u\n", clear_bit);
unsigned int toggle_bit = num ^ (1 << bit_position);
printf("Number with bit toggled: %u\n", toggle_bit);
int bit_status = (num >> bit_position) & 1;
printf("Bit status (0 or 1): %d\n", bit_status);
return 0;
}