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 sumDigitsIterative(int n) {
int sum = 0;
while (n > 0) {
sum += n % 10;
n /= 10;
}
return sum;
}
int sumDigitsRecursive(int n) {
if (n == 0) {
return 0;
}
return (n % 10) + sumDigitsRecursive(n / 10);
}
int main() {
int num = 12345;
printf("Iterative Sum of Digits: %d\n", sumDigitsIterative(num));
printf("Recursive Sum of Digits: %d\n", sumDigitsRecursive(num));
return 0;
}#include <stdio.h>
#include <stdbool.h>
bool isPalindrome(int num) {
int reversedNum = 0, originalNum = num, remainder;
while (num != 0) {
remainder = num % 10;
reversedNum = reversedNum * 10 + remainder;
num /= 10;
}
return originalNum == reversedNum;
}
int main() {
int number;
scanf("%d", &number);
if (isPalindrome(number)) {
printf("Palindrome");
} else {
printf("Not Palindrome");
}
return 0;
}#include <stdbool.h>
bool isPowerOfTwoIterative(unsigned int n) {
if (n == 0) return false;
while (n % 2 == 0) {
n /= 2;
}
return (n == 1);
}
bool isPowerOfTwoBitwise(unsigned int n) {
return (n != 0) && ((n & (n - 1)) == 0);
}#include <stdio.h>
long long power(long long base, long long exp, long long mod) {
long long res = 1;
base %= mod;
while (exp > 0) {
if (exp % 2 == 1)
res = (res * base) % mod;
base = (base * base) % mod;
exp /= 2;
}
return res;
}
int main() {
long long base = 2;
long long exp = 10;
long long mod = 1000000007;
long long result = power(base, exp, mod);
printf("%lld\n", result);
return 0;
}#include <stdio.h>
int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
int gcd_array(int arr[], int n) {
int result = arr[0];
for (int i = 1; i < n; i++) {
result = gcd(result, arr[i]);
}
return result;
}
int main() {
int arr[] = {12, 18, 24, 30};
int n = sizeof(arr) / sizeof(arr[0]);
int result = gcd_array(arr, n);
printf("GCD of the array is: %d\n", result);
return 0;
}#include <stdio.h>
int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
int main() {
int num1, num2;
scanf("%d %d", &num1, &num2);
printf("%d\n", lcm(num1, num2));
return 0;
}#include <stdio.h>
int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
int main() {
int num1, num2;
printf("Enter two integers: ");
scanf("%d %d", &num1, &num2);
printf("GCD of %d and %d is %d\n", num1, num2, gcd(num1, num2));
return 0;
}#include <stdio.h>
int factorial(int n) {
if (n == 0)
return 1;
else
return n * factorial(n - 1);
}
int isStrong(int num) {
int sum = 0, temp = num, digit;
while (temp > 0) {
digit = temp % 10;
sum += factorial(digit);
temp /= 10;
}
return (sum == num);
}
int main() {
int number;
scanf("%d", &number);
if (isStrong(number))
printf("Strong Number\n");
else
printf("Not Strong Number\n");
return 0;
}#include <stdio.h>
int countDigits(int n) {
int count = 0;
if (n == 0) return 1;
while (n > 0) {
n = n - (n / 10) * 10;
n = n / 10;
count++;
}
return count;
}
int main() {
int num = 12345;
int digitCount = countDigits(num);
printf("Number of digits in %d is %d\n", num, digitCount);
return 0;
}#include <stdio.h>
#include <stdbool.h>
#include <math.h>
bool isPrime(int num) {
if (num <= 1) return false;
for (int i = 2; i <= sqrt(num); i++) {
if (num % i == 0) return false;
}
return true;
}
void printPrimesInRange(int start, int end) {
for (int i = start; i <= end; i++) {
if (isPrime(i)) {
printf("%d ", i);
}
}
printf("\n");
}
int main() {
int start, end;
printf("Enter the start of the range: ");
scanf("%d", &start);
printf("Enter the end of the range: ");
scanf("%d", &end);
printf("Prime numbers in the range %d to %d are: ", start, end);
printPrimesInRange(start, end);
return 0;
}