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Formatted question description: https://leetcode.ca/all/717.html

# 717. 1-bit and 2-bit Characters (Easy)

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.

• class Solution {
public boolean isOneBitCharacter(int[] bits) {
int length = bits.length;
int index = 0;
while (index < length) {
if (index == length - 1)
return true;
int bit = bits[index];
if (bit == 0)
index++;
else if (bit == 1)
index += 2;
}
return false;
}
}

############

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

• // OJ: https://leetcode.com/problems/1-bit-and-2-bit-characters/
// 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;
}
};

• class Solution:
def isOneBitCharacter(self, bits: List[int]) -> bool:
i, n = 0, len(bits)
while i < n - 1:
i += bits[i] + 1
return i == n - 1

############

"""
class Solution(object):
def isOneBitCharacter(self, bits):
"""
:type bits: List[int]
:rtype: bool
"""
pos = 0
while pos < len(bits) - 1:
if bits[pos] == 1:
pos += 2
else:
pos += 1
return pos == len(bits) - 1 and bits[pos] == 0

• func isOneBitCharacter(bits []int) bool {
i, n := 0, len(bits)
for i < n-1 {
i += bits[i] + 1
}
return i == n-1
}

• /**
* @param {number[]} bits
* @return {boolean}
*/
var isOneBitCharacter = function (bits) {
let i = 0;
const n = bits.length;
while (i < n - 1) {
i += bits[i] + 1;
}
return i == n - 1;
};