## Solution 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:TrueExplanation: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:FalseExplanation: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`

.

*SOLUTION:*

```
auto speedup = []() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
return nullptr;
}();
class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
int i = bits.size() - 2;
while (i >= 0 && bits[i] > 0) i--;
return (bits.size() - i) % 2 == 0;
}
};
```