717. 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).

Given a binary array bits that ends with 0, return true if the last character must be a one-bit character.

 

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.

 

Constraints:

  • 1 <= bits.length <= 1000
  • bits[i] is either 0 or 1.

Solutions

Solution 1

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int i = 0, n = bits.size();
        while (i < n - 1) i += bits[i] + 1;
        return i == n - 1;
    }
};

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

JavaScript Code
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/**
 * @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;
};