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
    uint8_t cou=0;
    for(int i =0; i<=31; i++){
        if(n&(0x01<<i)){
            cou++;
        }
        if (cou==2){
            return "NO";
        }
    }
    if(cou==0){
        return "NO";
    }
    return "YES";
}

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

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

Solving Approach

Approach 1: just iterate through the integrer and if count value goes to 2 then return FALSE, however if the vale is 0 the cou value is 0 and 0 is not the power of 0 so just check explicitly for it in the end of the loop and then leave it.

Approach 2: Better one

    if (n > 0 && (n & (n - 1)) == 0)
        return "YES";
    else
        return "NO";