Sure, here is a step-by-step guide to solve this problem:

1. First, we need to understand what it means for a number to be a power of 2. A number is a power of 2 if it can be expressed as 2^n for some non-negative integer n. For example, 2, 4, 8, 16 are all powers of 2 because they can be expressed as 2^1, 2^2, 2^3, 2^4 respectively.

2. A simple way to check if a number is a power of 2 is to keep dividing the number by 2 until we get a number that is not divisible by 2. If that number is 1, then the original number is a power of 2. If it's anything other than 1, then the original number is not a power of 2.

3. Here is a simple algorithm to implement this:

- If N is 0, print 'NO' because 0 is not a power of 2.
- While N is greater than 1:
- If N is not divisible by 2, print 'NO' and stop.
- Divide N by 2.
- If we've exited the loop, print 'YES' because N is a power of 2.

4. This algorithm works because any number that is a power of 2 will eventually be reduced to 1 by repeatedly dividing by 2. Any number that is not a power of 2 will eventually produce a remainder when divided by 2.

5. This algorithm runs in O(log N) time because in the worst case, we have to divide N by 2 up to log N times.

Question

Sure, here is a step-by-step guide to solve this problem:

1. First, we need to understand what it means for a number to be a power of 2. A number is a power of 2 if it can be expressed as 2^n for some non-negative integer n. For example, 2, 4, 8, 16 are all powers of 2 because they can be expressed as 2^1, 2^2, 2^3, 2^4 respectively.

2. A simple way to check if a number is a power of 2 is to keep dividing the number by 2 until we get a number that is not divisible by 2. If that number is 1, then the original number is a power of 2. If it's anything other than 1, then the original number is not a power of 2.

3. Here is a simple algorithm to implement this:

- If N is 0, print 'NO' because 0 is not a power of 2.
   - While N is greater than 1:
     - If N is not divisible by 2, print 'NO' and stop.
     - Divide N by 2.
   - If we've exited the loop, print 'YES' because N is a power of 2.

4. This algorithm works because any number that is a power of 2 will eventually be reduced to 1 by repeatedly dividing by 2. Any number that is not a power of 2 will eventually produce a remainder when divided by 2.

5. This algorithm runs in O(log N) time because in the worst case, we have to divide N by 2 up to log N times.

Knowee AI · Accepted Answer

Sure, here is a step-by-step guide to solve this problem:

1. First, we need to understand what it means for a number to be a power of 2. A number is a power of 2 if it can be expressed as 2^n for some non-negative integer n. For example, 2, 4, 8, 16 are all powers of 2 because they can be expressed as 2^1, 2^2, 2^3, 2^4 respectively.

2. A simple way to check if a number is a power of 2 is to keep dividing the number by 2 until we get a number that is not divisible by 2. If that number is 1, then the original number is a power of 2. If it's anything other than 1, then the original number is not a power of 2.

3. Here is a simple algorithm to implement this:

- If N is 0, print 'NO' because 0 is not a power of 2.
   - While N is greater than 1:
     - If N is not divisible by 2, print 'NO' and stop.
     - Divide N by 2.
   - If we've exited the loop, print 'YES' because N is a power of 2.

4. This algorithm works because any number that is a power of 2 will eventually be reduced to 1 by repeatedly dividing by 2. Any number that is not a power of 2 will eventually produce a remainder when divided by 2.

5. This algorithm runs in O(log N) time because in the worst case, we have to divide N by 2 up to log N times.

Given a non-negative integer N, the task is to check if N is a power of 2. Print 'YES' if N is the power of 2; else, print 'NO'.

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