🚀 **Code Challenge
#115: Mastering Pointers in C!**

Hey Coders! 👩‍💻👨‍💻

Let's dive into the world of pointers with today's challenge! 🎯


#include <stdio.h>

int main() {
// Declare an integer variable
int num = 42;

// Declare a pointer and point it to the address of 'num'
int *ptr = &num;

// Modify the value using the pointer
*ptr += 8;

// Print the updated value
printf("The new value of num is: %d\n", num);

return 0;
}


🤔 Challenge Task:
1. Understand how the pointer ptr is initialized.
2. Use the pointer to increment the value of num by 8.
3. Print the updated value of num.

💡 Hint: Remember to dereference the pointer using * to access and modify the value it points to.

Happy coding! 🚀💻 #CodeChallenge #CProgramming #PointersInC
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🚀 Code Challenge #115: Array Manipulation in C!

Hello Coders! 👩‍💻👨‍💻

Let's continue our coding journey with today's challenge – Manipulating Arrays! 🌐🔍

116: Remove Duplicates from Array


#include<stdio.h>
#include<conio.h>

void main()
{
int array[10];
int i, n, pos, element, found = 0;
clrscr();

printf("Enter total number of elements : ");
scanf("%d", &n);
printf("\nEnter the elements : \n");

for (i = 0; i < n; i++)
{
scanf("%d", &array[i]);
}

printf("Input array elements are\n");

for (i = 0; i < n; i++)
{
printf("\n%d", array[i]);
}

printf("\nEnter the element to be deleted : ");

scanf("%d", &element);

for (i = 0; i < n; i++)
{
if (array[i] == element)
{
found = 1;
pos = i;
break;
}
}

if (found == 1)
{
for (i = pos; i < n - 1; i++)
{
array[i] = array[i + 1];
}

printf("\nResultant array elements are : ");

for (i = 0; i < n - 1; i++)
{
printf("\n%d", array[i]);
}
}
else
{
printf("\nElement %d is not found in the array", element);
}
getch();
}


🤔 Challenge Task:
Modify the provided code to remove duplicates from the array and display the result.

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Happy coding! 🚀💻 #CodeChallenge #CProgramming #ArrayManipulation

🚀 Code Challenge #116: Searching in a Sorted Array

Hello, Coding Enthusiasts! 👩‍💻👨‍💻

Dive into today's challenge focused on searching in a sorted array! 🔍🌐

Question 116: Binary Search in a Sorted Array


#include <stdio.h>

int binarySearch(int array[], int low, int high, int target) {
while (low <= high) {
int mid = low + (high - low) / 2;

if (array[mid] == target)
return mid;

if (array[mid] < target)
low = mid + 1;
else
high = mid - 1;
}

return -1; // Target not found
}

int main() {
int array[] = {2, 5, 8, 12, 16, 23, 38, 42, 56, 72};
int n = sizeof(array) / sizeof(array[0]);
int target = 23;

int result = binarySearch(array, 0, n - 1, target);

if (result != -1)
printf("Element %d found at index %d", target, result);
else
printf("Element %d not found in the array");

return 0;
}

🤔 Challenge Task:
1. Understand the provided binary search algorithm.
2. Execute the algorithm to find the index of the target element (e.g., 23) in the sorted array.

Happy coding! 🚀💻
Join us on [Telegram](https://t.me/programming_and_coding) for more coding discussions and challenges! #CodeChallenge #CProgramming #BinarySearch

🚀 Coding Wisdom #120: The Art of Problem Solving

Hello, Coding Maestros! 👩‍💻👨‍💻

Embark on a journey of problem-solving, where the code becomes a canvas and you, the artist! 🎨✨

🔍 The Problem-Solving Mindset:
1. Understand the Problem:
- Break it down into smaller parts.
- Clearly define the input, output, and constraints.

2. Explore Strategies:
- Consider different approaches.
- Think about time and space complexity.

3. Start with Pseudocode:
- Draft a high-level plan before diving into code.
- Ensure your logic aligns with the problem.

🌐 Unique Code Challenge: Rotate Array Elements

#include <stdio.h>

void rotateArray(int arr[], int n, int k) {
    for (int i = 0; i < k; i++) {
        int temp = arr[0];
        for (int j = 0; j < n - 1; j++) {
            arr[j] = arr[j + 1];
        }
        arr[n - 1] = temp;
    }
}

int main() {
    int array[] = {1, 2, 3, 4, 5};
    int n = sizeof(array) / sizeof(array[0]);
    int k = 2;

    rotateArray(array, n, k);

    printf("Array after rotating %d times:\n", k);
    for (int i = 0; i < n; i++) {
        printf("%d ", array[i]);
    }

    return 0;
}

💡 Insight:
Explore a unique solution to rotating array elements. Understand the logic and adapt it to different scenarios.

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Let's elevate our coding wisdom together! 🚀💬 #ProblemSolving #CodingWisdom #CodeChallenge