Leetcode: 1-bit and 2-bit Characters

1-bit and 2-bit Characters



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.

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/1-bit-and-2-bit-characters
## Basic Ideas: One pass
##       If current digit is 0, move 1 step
##       If current digit is 1, move 2 steps
##       Check whether the last match is 1 step
##       Note: since the last digit is 0, the given string can always match
## Complexity: Time O(n), Space O(1)
class Solution(object):
    def isOneBitCharacter(self, bits):
        """
        :type bits: List[int]
        :rtype: bool
        """
        length = len(bits)
        if length == 0:
            return False
        if length == 1:
            return bits[0] == 0

        i = 0
        while i< length:
            if bits[i] == 0:
                i += 1
            else:
                i += 2
            if i == length - 1:
                return True
        return False
linkedin
github
slack

Share It, If You Like It.

Leave a Reply

Your email address will not be published.