## Description

We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11). Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.

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Example 1: Input: bits = [1, 0, 0] Output: True Explanation: The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character. Example 2: Input: bits = [1, 1, 1, 0] Output: False Explanation: The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character. Note: 1 <= len(bits) <= 1000. bits[i] is always 0 or 1. |

## Main Idea

We can directly tell the answer in 3 cases: [1, 0], [1, 1] ,and [0]. Hence we can use the recursion solution to turn the bits to these cases. If bits start with number 0, then we decode it’s a 1-bit character, it’s a 2-bit character otherwise.

## Solution

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class Solution: def isOneBitCharacter(self, bits) -> bool: if bits == [0]: return True if bits == [1,0] or bits == [1,1]: return False if bits[0] == 0: return self.isOneBitCharacter(bits[1:]) if bits[0] == 1: return self.isOneBitCharacter(bits[2:]) |