Skip to main content

Featured

C Program to Solve Two Sum Using Brute Force (With Algorithm & Output)

 Introduction The Two Sum problem is a popular coding interview question where we must find two indices of an array whose values add up to a given target. This program demonstrates a simple brute-force solution in C using nested loops and dynamic memory allocation. Problem Statement Given an integer array and a target value, return the indices of the two numbers such that they add up to the target. Each input has exactly one solution, and the same element cannot be used twice. The result should return the indices, not the values. If no solution exists, return NULL.  Algorithm / Logic Explanation Start the program. Traverse the array using a loop from index 0 to numsSize - 1 . Inside this loop, use another loop starting from i + 1 to numsSize - 1 . For every pair (i, j) , check if nums[i] + nums[j] == target . If condition becomes true: Allocate memory for 2 integers using malloc() . Store indices i and j . Set returnSize = 2 . Return the result poi...
Post a Comment
Read more
Labels

C program to check whether a given number is a power of 2 using bitwise operations

Check Power of 2 in C

⚡ Check If a Number Is a Power of 2

#include<stdio.h>

int power(int num)
{
    if(num > 0 && (num & (num - 1)) == 0)
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

int main()
{
    int num;
    scanf("%d", &num);
    
    if(power(num))
    {
        printf("%d is a power of 2\n", num);
    }
    else
    {
        printf("%d is not a power of 2\n", num);
    }
}

📘 Explanation:

✅ The condition (num & (num - 1)) == 0 is true only if there is only one set bit in num.
✅ This logic efficiently checks whether a number is a power of 2.
✅ It excludes 0 and negative numbers.

🧪 Sample Output:

Input:
8
Output:
8 is a power of 2

🏷️ Keywords:

C program power of 2, bitwise check power of 2, efficient logic, beginner C programming, technical interview questions

Labels:

Comments

Post a Comment

Popular Posts

🌙