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
    if(n!=0 &((n&(n-1))==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

it was a trick question , n&(n-1) clears the least set bit in n, if n is a power of 2, only 1 bit is set ,so the answer is 0.