Check if a number has bits in alternate pattern | Set-2 O(1) Approach

Given a positive integer n. The problem is to check whether this integer has an alternate pattern in its binary representation or not. Here alternate pattern means that the set and unset bits in n occur in alternate order. For example- 5 has an alternate pattern i.e. 101. 
Print “Yes” if it has an alternate pattern otherwise “No”. 
Note: 0 < n.
Examples : 
 

Input : 10
Output : Yes
(10)10 = (1010)2, has an alternate pattern.

Input : 12
Output : No
(12)10 = (1100)2, does not have an alternate pattern.

Simple Approach: It has been discussed in this post having a time complexity of O(n).
Efficient Approach: Following are the steps:
 

1. Calculate num = n ^ (n >> 1). If n has an alternate pattern, then n ^ (n >> 1) operation will produce a number having set bits only. '^' is a bitwise XOR operation.
2. Check whether all the bits in num are set or not. Refer this post.

Output : 
 

Yes

Time Complexity : O(1)

Auxiliary Space : O(1)