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231 Power of Two - Easy

Problem:

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?

Problem Analysis:

  • use recursion or
  • use log or
  • use bit manipulation
    • (n & (n - 1)) == 0: This part is the key to checking whether n is a power of two.
    • n - 1 subtracts 1 from n. This operation flips the rightmost set bit in the binary representation of n to 0 while leaving all other bits the same.
    • n & (n - 1) performs a bitwise AND operation between n and n - 1. In this operation, the result will be zero if and only if n has exactly one bit set to 1 in its binary representation. This is because when you AND a number with itself minus one, all the bits to the right of the rightmost set bit become zero, and all other bits remain unchanged.

Solutions:

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        return n>0 and (n & (n - 1)) == 0

or

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        return n > 0 and int(math.log10(n) / math.log10(2)) == (math.log10(n) / math.log10(2))

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