231. Power of Two
Difficulty: Easy
Topics: Math Bit Manipulation Recursion
Statement: Given an integer n, return true if it is a power of two, and false otherwise. An integer n is a power of two if there exists an integer x greater than or equal to 0 such that n == 2 raised to x.
Example 1: Input n = 1 Output true Explanation 2 raised to 0 = 1
Example 2: Input n = 16 Output true Explanation 2 raised to 4 = 16
Example 3: Input n = 3 Output false
Constraints: -2 raised to 31 <= n <= 2 raised to 31 minus 1
Follow up: Can you solve it without loops or recursion?
Summarized solution: You need to check that n is positive and that its binary representation has only one bit set to 1. For any n that is a power of two, the bitwise operation n AND n minus 1 returns zero.
Detailed approach: 1 Check that n > 0, otherwise return false 2 Apply the bitwise operation between n and n-1. If n AND (n-1) == 0, then n is a power of two and we return true; otherwise, we return false
Example implementation in pseudocode: function isPowerOfTwo(n) { if n <= 0 return false; return (n AND (n-1)) == 0; }
Complexity: Time O(1) Space O(1)
Test cases: isPowerOfTwo(1) returns true isPowerOfTwo(16) returns true isPowerOfTwo(3) returns false isPowerOfTwo(0) returns false isPowerOfTwo(-2) returns false
Final explanation: The check is safe for the 32-bit integer range and avoids loops or recursion by using a very efficient bitwise operation. Any number that is not positive or has more than one active bit is not a power of two.
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