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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.
  • Related Topics:
    Array

    Similar Questions:

    Solution 1.

    // OJ: https://leetcode.com/problems/1-bit-and-2-bit-characters/
    // Author: github.com/lzl124631x
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        bool isOneBitCharacter(vector<int>& A) {
            for (int i = 0, N = A.size(); i < N; ) {
                if (A[i] == 1) {
                    if (i + 2 < N) i += 2;
                    else return false;
                } else if (i == N - 1) return true;
                else ++i;
            }
            return false;
        }
    };