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C Programming Language || Hands On Coding

C Programming Language || Hands On Coding

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Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii

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📈 Analytical overview of Telegram channel C Programming Language || Hands On Coding

Channel C Programming Language || Hands On Coding (@c_programming_language_coding) in the English language segment is an active participant. Currently, the community unites 12 799 subscribers, ranking 9 568 in the Technologies & Applications category and 30 934 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 12 799 subscribers.

According to the latest data from 30 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -216 over the last 30 days and by -11 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 7.69%. Within the first 24 hours after publication, content typically collects 2.42% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 985 views. Within the first day, a publication typically gains 310 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • 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:
Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii

Thanks to the high frequency of updates (latest data received on 31 August, 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.

12 799
Subscribers
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Posts Archive
#Cprogramming #Palindrome #InterviewQuestion

Check Palindrome Number
#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;
}

#CProgramming #BitwiseOperators #PowerOfTwo

Check Power of 2 (Iterative and Bitwise)
#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);
}

#CProgramming #Algorithms #Exponentiation

Fast Exponentiation (Binary Exponentiation)
#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;
}

#CProgramming #GCD #Arrays

GCD of Array Elements
#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;
}

#CProgramming #LCM #GCD

LCM of Two Numbers
#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;
}

#CProgramming #GCD #EuclideanAlgorithm

GCD of Two Numbers (Euclidean Algorithm)
#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;
}

#CProgramming #StrongNumber #InterviewPrep

Check Strong Number
#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;
}

#CProgramming #DigitsCount #InterviewPrep

Count Digits Without % or /
#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;
}

#CProgramming #PrimeNumbers

Check Prime Number in Range
#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;
}

#CProgramming #DigitsCount #InterviewPrep

Count Digits Without % or /
#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;
}