## 693. Binary Number with Alternating Bits

Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101


Example 2:

Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.


Example 3:

Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.


Example 4:

Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.


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#### Approach #1: Convert to String [Accepted]

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Intuition and Algorithm

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Let\'s convert the given number into a string of binary digits. Then, we should simply check that no two adjacent digits are the same.

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Complexity Analysis

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Time Complexity: . For arbitrary inputs, we do work, where is the number of bits in n. However, .

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Space complexity: , or alternatively .

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#### Approach #2: Divide By Two [Accepted]

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Intuition and Algorithm

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We can get the last bit and the rest of the bits via n % 2 and n // 2 operations. Let\'s remember cur, the last bit of n. If the last bit ever equals the last bit of the remaining, then two adjacent bits have the same value, and the answer is False. Otherwise, the answer is True.

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Also note that instead of n % 2 and n // 2, we could have used operators n & 1 and n >>= 1 instead.

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Complexity Analysis

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Time Complexity: . For arbitrary inputs, we do work, where is the number of bits in n. However, .

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Space complexity: .

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Analysis written by: @awice

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