LeetCode 717 - 1-bit and 2-bit Characters

https://leetcode.com/problems/1-bit-and-2-bit-characters/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.

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.

When reading from the i-th position, if bits[i] == 0, the next character must have 1 bit; else if bits[i] == 1, the next character must have 2 bits. We increment our read-pointer i to the start of the next character appropriately. At the end, if our pointer is at bits.length - 1, then the last character must have a size of 1 bit.

    public boolean isOneBitCharacter(int[] bits) {
        int i = 0;
        while (i < bits.length - 1) {
            i += bits[i] + 1;
        }
        return i == bits.length - 1;
    }

https://leetcode.com/problems/1-bit-and-2-bit-characters/discuss/108969/Java-solution-1-or-2

    public boolean isOneBitCharacter(int[] bits) {
        int n = bits.length, i = 0;
        while (i < n - 1) {
            if (bits[i] == 0) i++;
            else i += 2;
        }
        return i == n - 1;
    }

The second-last 0 must be the end of a character (or, the beginning of the array if it doesn't exist). Looking from that position forward, the array bits takes the form [1, 1, ..., 1, 0] where there are zero or more 1's present in total. It is easy to show that the answer is true if and only if there are an even number of ones present.

In our algorithm, we will find the second last zero by performing a linear scan from the right. We present two slightly different approaches below.

    public boolean isOneBitCharacter(int[] bits) {
        int i = bits.length - 2;
        while (i >= 0 && bits[i] > 0) i--;
        return (bits.length - i) % 2 == 0;
    }

https://leetcode.com/problems/1-bit-and-2-bit-characters/discuss/108967/JAVA-check-only-the-end-of-array

We don't need to traverse the whole array, just check the last part of it.

1. if there is only one symbol in array the answer is always true (as last element is 0)
2. if there are two 0s at the end again the answer is true no matter what the rest symbols are( ...1100, ...1000,)
3. if there is 1 right before the last element(...10), the outcome depends on the count of sequential 1, i.e.
   a) if there is odd amount of 1(10, ...01110, etc) the answer is false as there is a single 1 without pair
   b) if it's even (110, ...011110, etc) the answer is true, as 0 at the end doesn't have anything to pair with

    public boolean isOneBitCharacter(int[] bits) {
        int ones = 0;
        //Starting from one but last, as last one is always 0.
        for (int i = bits.length - 2; i >= 0 && bits[i] != 0 ; i--) { 
            ones++;
        }
        if (ones % 2 > 0) return false; 
        return true;
    }