Check If a Number Is a Power of Two

Code

#include <stdio.h>
#include <stdint.h>

// Complete the function
const char* is_power_of_two(uint32_t n) {
    // Your logic here
    uint32_t m;
    m=n-1;

    if ((n >0) && ((n & m) == 0)) 
        return "YES";
    else
        return "NO";
}

int main() {
    uint32_t n;
    scanf("%u", &n);

    const char* result = is_power_of_two(n);
    printf("%s", result);
    return 0;
}

Solving Approach

the number has to be positive and bitwise AND with (n-1) gotta be zero

If you subtract 1 from a power of two, all the bits to the right of the leading '1' become '1', and the leading '1' becomes '0'.