dsa-competitive-programming/leetcode/lc693/AlternatingXOR.cpp

class Solution{
    public : 
        bool hasAlternatingBits(int n){
            int m = n^(n>>1);
            return (m&(m+1)==0);
        }
}

// TC & SC : o(1)
// n>>1 inverts all bits and XOR makes m have all set bits if n had alternating bits. (m+1) makes all bits of m as zero and m&(m+1) will be zero