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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 12 318 subscribers, ranking 10 235 in the Technologies & Applications category and 31 257 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 12 318 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.

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 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.

12 318
Subscribers
-924 hours
-837 days
-23630 days
Posts Archive
#CPP #BitManipulation #BrianKernighan

Count Set Bits (Brian Kernighan's Algorithm)
#include <iostream>

using namespace std;

int countSetBits(int n) {
    int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}

int main() {
    int num = 7; 
    cout << "Number of set bits in " << num << " is " << countSetBits(num) << endl;
    return 0;
}

#CProgramming #BitwiseOperators #SwapTechnique

Swap Two Numbers using XOR
#include <stdio.h>

void swap(int *x, int *y) {
    *x = *x ^ *y;
    *y = *x ^ *y;
    *x = *x ^ *y;
}

int main() {
    int a = 10, b = 5;
    swap(&a, &b);
    printf("a = %d, b = %d\n", a, b);
    return 0;
}

#CProgramming #Combinations #nCr

Find nCr (Combinations)
#include <stdio.h>

long long factorial(int n) {
    if (n == 0)
        return 1;
    else
        return n * factorial(n - 1);
}

long long nCr(int n, int r) {
    if (r < 0 || r > n)
        return 0;
    if (r == 0 || r == n)
        return 1;
    if (r > n / 2)
        r = n - r;
    return factorial(n) / (factorial(r) * factorial(n - r));
}

int main() {
    int n, r;
    printf("Enter n and r: ");
    scanf("%d %d", &n, &r);
    printf("nCr = %lld\n", nCr(n, r));
    return 0;
}

#CProgramming #PascalTriangle #InterviewPrep

Print Pascal's Triangle
#include <stdio.h>

int main() {
  int rows, coef = 1, space, i, j;

  scanf("%d", &rows);

  for (i = 0; i < rows; i++) {
    for (space = 1; space <= rows - i; space++)
      printf("  ");

    for (j = 0; j <= i; j++) {
      if (j == 0 || i == 0)
        coef = 1;
      else
        coef = coef * (i - j + 1) / j;

      printf("%4d", coef);
    }
    printf("\n");
  }

  return 0;
}

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#Cprogramming #InterviewPrep #SumOfNumbers

Sum of First N Even/Odd Numbers
#include <stdio.h>

int sumOfFirstNEven(int n) {
    return n * (n + 1);
}

int sumOfFirstNOdd(int n) {
    return n * n;
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", &n);
    printf("Sum of first %d even numbers: %d\n", n, sumOfFirstNEven(n));
    printf("Sum of first %d odd numbers: %d\n", n, sumOfFirstNOdd(n));
    return 0;
}

#CProgramming #Palindrome #PrimeNumbers

Check if Number is Palindromic Prime
#include <stdio.h>
#include <stdbool.h>
#include <math.h>

bool isPalindrome(int n) {
    int reversed = 0, original = n, remainder;
    while (n != 0) {
        remainder = n % 10;
        reversed = reversed * 10 + remainder;
        n /= 10;
    }
    return original == reversed;
}

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) return false;
    }
    return true;
}

bool isPalindromicPrime(int n) {
    return isPalindrome(n) && isPrime(n);
}

int main() {
    int num;
    scanf("%d", &num);
    if (isPalindromicPrime(num)) {
        printf("True\n");
    } else {
        printf("False\n");
    }
    return 0;
}

#CProgramming #Factorial #TrailingZeroes

Count Trailing Zeroes in Factorial
#include <stdio.h>

int countTrailingZeroes(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) {
    count += n / i;
  }
  return count;
}

int main() {
  int num = 25;
  printf("Trailing zeroes in %d! = %d\n", num, countTrailingZeroes(num));
  return 0;
}

#CProgramming #BinaryToDecimal #InterviewPrep

Convert Binary to Decimal
#include <stdio.h>
#include <math.h>

int binaryToDecimal(long long n) {
    int decimalNumber = 0, i = 0, remainder;
    while (n!=0) {
        remainder = n%10;
        n /= 10;
        decimalNumber += remainder*pow(2,i);
        ++i;
    }
    return decimalNumber;
}

int main() {
    long long n;
    printf("Enter a binary number: ");
    scanf("%lld", &n);
    printf("Decimal equivalent: %d", binaryToDecimal(n));
    return 0;
}