LeetCode 0717. 1-bit and 2-bit Characters Solution in Java, C++, Python & More | Explanation + Code

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 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 true; otherwise, return false.

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

PythonJavaC++GoTypeScriptRustJavaScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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

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

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 } }(code-box)

/** * @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; };(code-box)