Jobs

Interviews

Discussions

1-bit and 2-bit Characters

Easy

Quora

0 views

Topics:

Arrays

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.

Try coding it in LeetCode

1-bit and 2-bit Characters

Easy

Quora

0 views

Topics:

Arrays

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input array `bits` be empty or null?
What is the maximum size of the input array `bits`?
Is it guaranteed that the input array `bits` will always represent a valid sequence of 1-bit and 2-bit characters, or could there be invalid sequences (e.g., ending with '1')?
If the input array contains only one bit, what should the output be?
Are the bits in the input array guaranteed to be either 0 or 1?

Brute Force Solution

Approach

The brute force approach to this problem is like trying every single possible way to decode a secret message. We explore each path, one at a time, to see if it leads us to the correct conclusion.

Here's how the algorithm would work step-by-step:

Start at the beginning of the message.
Consider the first character as a possible 1-bit character and see where that leads.
Then, consider the first two characters as a possible 2-bit character and see where that leads.
For each of these possibilities, repeat the process for the remaining characters.
Continue this process until you reach the end of the message.
If, after considering all possibilities, at least one path ends with a 1-bit character as the last character in the message, then the answer is yes.
If no path ends with the 1-bit character in the last position, the answer is no.

Code Implementation

def is_last_character_valid(bits):
    number_of_bits = len(bits)
    
    def solve(current_index):
        # Base case: Reached the end of the array
        if current_index == number_of_bits:
            return True

        # If we've gone past the end, it's an invalid path
        if current_index > number_of_bits:
            return False

        # Try one-bit character
        if bits[current_index] == 0:
            if solve(current_index + 1):
                return True

        # Try two-bit character
        if current_index + 1 < number_of_bits:
            # Need to check bounds before accessing bits[current_index + 1]
            if bits[current_index] == 1:
                if solve(current_index + 2):
                    return True

        return False

    # Initiate recursion to explore all combinations
    return solve(0)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible combination of 1-bit and 2-bit characters. At each position, we have two choices: consider it a 1-bit character or part of a 2-bit character. This branching behavior leads to a binary tree-like exploration of the input array of size n. Therefore, the number of possible paths grows exponentially with the input size, resulting in O(2^n) time complexity as we might explore all 2 to the power of n branches.

Space Complexity

O(N) – The provided plain English explanation outlines a brute-force approach that implicitly uses recursion to explore all possible decoding paths. Each recursive call represents a choice between considering a 1-bit or a 2-bit character. In the worst-case scenario, the depth of the recursion can reach N, where N is the number of bits in the input array, bits. Each level of recursion adds a new frame to the call stack. Thus, the auxiliary space used by the recursion stack grows linearly with the input size, resulting in a space complexity of O(N).

Optimal Solution

Approach

The key is to walk through the given sequence of bits and determine the last character type directly. Since we only care about the last character, we can skip ahead when we encounter a two-bit character to efficiently determine if the last character is a one-bit character.

Here's how the algorithm would work step-by-step:

Begin at the very start of the sequence.
Check each bit in the sequence, moving forward one bit at a time.
If you find a one-bit character (a '0'), just move to the next bit.
If you find a two-bit character (a '1'), you know it takes up the current bit AND the next bit, so skip ahead *two* bits.
Keep doing this until you reach the end of the sequence.
If, when you get to the end, you are standing on the last bit, then the last character must have been a one-bit character (a '0').
If you skipped over the last bit (because you encountered a '1' and jumped ahead two spaces) then the last character was not a one-bit character.

Code Implementation

def is_last_character_one_bit_character(bits):
    current_index = 0

    while current_index < len(bits):
        # If the current bit is 1, it's a two-bit character.
        if bits[current_index] == 1:
            current_index += 2

        # Otherwise it is a one-bit character.
        else:
            current_index += 1

    # Checks if it ends on the last index.
    return current_index == len(bits)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of bits once. In each iteration, it either advances by one position (when encountering a '0') or by two positions (when encountering a '1'). Since the algorithm only moves forward and never revisits a position, and it terminates when it reaches the end of the array of size n, the maximum number of iterations is proportional to n. Thus, the time complexity is O(n).

Space Complexity

O(1) – The algorithm iterates through the input array, but it only uses a single integer variable to track the current position. No additional data structures like arrays, lists, or hash maps are created. The space used by this integer variable remains constant regardless of the input array's size (N, the number of bits), so the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return false if the array is null or empty as there's no character to analyze.
Array with only one element: [0] or [1]	If the single element is 0, return true; if it's 1, return false because 1 must be the start of a 2-bit character but has no following bit.
Array ending with [1]	If the array ends with a single '1', it indicates an incomplete 2-bit character, making the last complete character not a 1-bit character, so return false.
Array ending with [1,0] or [1,1]	These are valid 2-bit characters followed by no more characters, indicating the last character is not a 1-bit character so return false.
Array of all zeros	The last character is always a 1-bit character, so return true.
Array starting with a long sequence of ones followed by a zero	Iterate through the array, skipping two positions for each '1' encountered, and return true if the final index reached is the second to last element.
Large input array close to memory limits	The iterative solution uses constant space, so it scales efficiently with large inputs.
Array consisting of repeating patterns like [1,0,1,0,1,0]	The standard iteration will correctly handle these patterns by advancing either one or two steps at a time.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input array `bits` be empty or null?
What is the maximum size of the input array `bits`?
Is it guaranteed that the input array `bits` will always represent a valid sequence of 1-bit and 2-bit characters, or could there be invalid sequences (e.g., ending with '1')?
If the input array contains only one bit, what should the output be?
Are the bits in the input array guaranteed to be either 0 or 1?

Brute Force Solution

Approach

The brute force approach to this problem is like trying every single possible way to decode a secret message. We explore each path, one at a time, to see if it leads us to the correct conclusion.

Here's how the algorithm would work step-by-step:

Start at the beginning of the message.
Consider the first character as a possible 1-bit character and see where that leads.
Then, consider the first two characters as a possible 2-bit character and see where that leads.
For each of these possibilities, repeat the process for the remaining characters.
Continue this process until you reach the end of the message.
If, after considering all possibilities, at least one path ends with a 1-bit character as the last character in the message, then the answer is yes.
If no path ends with the 1-bit character in the last position, the answer is no.

Code Implementation

def is_last_character_valid(bits):
    number_of_bits = len(bits)
    
    def solve(current_index):
        # Base case: Reached the end of the array
        if current_index == number_of_bits:
            return True

        # If we've gone past the end, it's an invalid path
        if current_index > number_of_bits:
            return False

        # Try one-bit character
        if bits[current_index] == 0:
            if solve(current_index + 1):
                return True

        # Try two-bit character
        if current_index + 1 < number_of_bits:
            # Need to check bounds before accessing bits[current_index + 1]
            if bits[current_index] == 1:
                if solve(current_index + 2):
                    return True

        return False

    # Initiate recursion to explore all combinations
    return solve(0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Begin at the very start of the sequence.
Check each bit in the sequence, moving forward one bit at a time.
If you find a one-bit character (a '0'), just move to the next bit.
If you find a two-bit character (a '1'), you know it takes up the current bit AND the next bit, so skip ahead *two* bits.
Keep doing this until you reach the end of the sequence.
If, when you get to the end, you are standing on the last bit, then the last character must have been a one-bit character (a '0').
If you skipped over the last bit (because you encountered a '1' and jumped ahead two spaces) then the last character was not a one-bit character.

Code Implementation

def is_last_character_one_bit_character(bits):
    current_index = 0

    while current_index < len(bits):
        # If the current bit is 1, it's a two-bit character.
        if bits[current_index] == 1:
            current_index += 2

        # Otherwise it is a one-bit character.
        else:
            current_index += 1

    # Checks if it ends on the last index.
    return current_index == len(bits)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return false if the array is null or empty as there's no character to analyze.
Array with only one element: [0] or [1]	If the single element is 0, return true; if it's 1, return false because 1 must be the start of a 2-bit character but has no following bit.
Array ending with [1]	If the array ends with a single '1', it indicates an incomplete 2-bit character, making the last complete character not a 1-bit character, so return false.
Array ending with [1,0] or [1,1]	These are valid 2-bit characters followed by no more characters, indicating the last character is not a 1-bit character so return false.
Array of all zeros	The last character is always a 1-bit character, so return true.
Array starting with a long sequence of ones followed by a zero	Iterate through the array, skipping two positions for each '1' encountered, and return true if the final index reached is the second to last element.
Large input array close to memory limits	The iterative solution uses constant space, so it scales efficiently with large inputs.
Array consisting of repeating patterns like [1,0,1,0,1,0]	The standard iteration will correctly handle these patterns by advancing either one or two steps at a time.