#717

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

1-bit and 2-bit Characters

Determine whether the last character in a binary array represents a one-bit character or a two-bit character.

Problem Statement

You are given a binary array bits that ends with a 0. Each element in the array can either be a 1 or a 0. The problem is to determine if the last character must be a one-bit character. Characters are either one-bit (represented by 0) or two-bit (represented by 10 or 11). You need to return true if the last character is a one-bit character, and false if it is not.

For example, consider the binary array [1, 0, 0]. This array can be decoded into two bits for the first two characters, leaving the last one as a one-bit character. On the other hand, for the binary array [1, 1, 1, 0], the last character cannot be a one-bit character, as the sequence represents two two-bit characters.

Examples

Example 1

Input: bits = [1,0,0]

Output: true

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

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.

Solution Approach

Track Position of the Next Character

To solve the problem, track the position of the next character by iterating over the array from left to right. Every time a 1 is encountered, skip the next bit (because it forms a two-bit character). If a 0 is encountered, move to the next bit as it forms a one-bit character. At the end of the loop, check whether you're at the last bit and decide if it is a one-bit character.

Efficient Array Traversal

The problem requires a single pass through the array, which ensures an O(N) time complexity. By checking each bit sequentially and adjusting the index accordingly (jumping over two bits after a 1), you minimize unnecessary iterations and keep space complexity constant at O(1). This solution ensures an efficient traversal of the array with minimal overhead.

Handle Edge Cases

Ensure that you handle edge cases like the smallest possible array with only one bit, [0], or arrays where all elements are 1's followed by 0. Properly handling these edge cases is crucial for ensuring correctness across all valid inputs, particularly with arrays near the constraint boundaries.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(1)

The time complexity of the solution is O(N) because we iterate through the array once. For each element, we only make a constant-time decision, either skipping one or two positions based on the current bit. The space complexity is O(1) since we only use a few variables to track the current position, regardless of the input size.

What Interviewers Usually Probe

Candidates who understand the importance of tracking the next character's position and the array-driven solution approach will likely solve the problem quickly.
Look for candidates who can explain the logic of skipping bits after encountering a 1 and how it ensures correctness.
Candidates who provide a detailed explanation of edge cases, such as handling arrays of length 1 or arrays with consecutive 1's followed by 0, will demonstrate thorough understanding.

Common Pitfalls or Variants

Common pitfalls

Candidates may forget to account for the scenario where the array ends with a 0, and incorrectly assume that the last bit is always a one-bit character.
Some candidates might miss the optimization of skipping two bits after a 1, leading to unnecessary iterations.
Candidates may overlook handling edge cases, like the smallest array or cases where there are multiple consecutive 1's followed by 0.

Follow-up variants

What if the array ends with a 1 instead of a 0? How would you handle that case?
Can this problem be extended to a more general case where bits represent a larger set of characters?
What if the input array is given in a different encoding format, like a string of 0's and 1's?

FAQ

What is the main challenge in the 1-bit and 2-bit Characters problem?

The main challenge lies in correctly identifying whether the last character in the binary array is a one-bit character or a two-bit character by decoding the array from left to right.

How do I track the current character in the array?

You track the current character by iterating through the array and adjusting the index. After a 1, skip the next bit to account for the two-bit character.

What is the time complexity of the solution?

The time complexity of the solution is O(N), where N is the length of the array. This is because we only traverse the array once.

Can this problem be solved with more advanced algorithms?

No, the problem is simple and can be efficiently solved with an O(N) time complexity. More complex algorithms aren't necessary for this problem.

How does GhostInterview help with solving this problem?

GhostInterview guides you step-by-step through the solution process, helping you optimize your approach and avoid common pitfalls like missing edge cases.

terminal

Solution

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:

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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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