In this guide, you will get 231. Power of Two LeetCode Solution with the best time and space complexity. The solution to Power of Two problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Given an integer n, return true if it is a power of two. Otherwise, return false.
An integer n is a power of two, if there exists an integer x such that n == 2x.
Example 1:
Input: n = 1
Output: true
Explanation: 20 = 1
Example 2:
Input: n = 16
Output: true
Explanation: 24 = 16
Example 3:
Input: n = 3
Output: false
Constraints:
-231 <= n <= 231 – 1
Follow up: Could you solve it without loops/recursion?
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(1)
231. Power of Two LeetCode Solution in C++
class Solution {
public:
bool isPowerOfTwo(int n) {
return n >= 0 && __builtin_popcount(n) == 1;
}
}; /* code provided by PROGIEZ */
231. Power of Two LeetCode Solution in Java
class Solution {
public boolean isPowerOfTwo(int n) {
return n >= 0 && Integer.bitCount(n) == 1;
}
} // code provided by PROGIEZ
231. Power of Two LeetCode Solution in Python
class Solution:
def isPowerOfTwo(self, n: int) -> bool:
return n >= 0 and n.bit_count() == 1 # code by PROGIEZ
