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: Direct Traversal

We can directly traverse the first \(n-1\) elements of the array \(\textit{bits}\), and each time decide how many elements to skip based on the value of the current element:

• If the current element is \(0\), skip \(1\) element (representing a one-bit character);
• If the current element is \(1\), skip \(2\) elements (representing a two-bit character).

When the traversal ends, if the current index equals \(n-1\), it means the last character is a one-bit character, and we return \(\text{true}\); otherwise, return \(\text{false}\).

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{bits}\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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function isOneBitCharacter(bits: number[]): boolean {
    let i = 0;
    const n = bits.length;
    while (i < n - 1) {
        i += bits[i] + 1;
    }
    return i === n - 1;
}

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impl Solution {
    pub fn is_one_bit_character(bits: Vec<i32>) -> bool {
        let mut i = 0usize;
        let n = bits.len();
        while i < n - 1 {
            i += (bits[i] + 1) as usize;
        }
        i == n - 1
    }
}

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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;
};

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