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In this approach, we will use bitwise operations to compare each bit with its adjacent one. The key is to repeatedly XOR the number with itself shifted one position to the right, which will result in a number with all bits set to 1 if the original number has alternating bits. Such a number should satisfy (n & (n + 1)) == 0.
Time Complexity: O(1), Space Complexity: O(1)
1public class Solution {
2 public boolean hasAlternatingBits(int n) {
3 int xor = n ^ (n >> 1);
4 return (xor & (xor + 1)) == 0;
5 }
6}The XOR operation is applied between the number and its right shift which will allow us to determine if it has alternating bits by checking all bits are 1s, using the given condition.
This approach involves checking each bit pair iteratively. We will repeatedly check if the last two bits are the same or different by analyzing the least significant bits and shifting the number to the right iteratively.
Time Complexity: O(log N), where N is the value of the number. Space Complexity: O(1)
1
This implementation analyzes the last two bits of the number by checking the modulus and right-shifting iteratively. It ensures they are different, ensuring the bits alternate.