How do you solve 1-bit and 2-bit Characters on LeetCode?

1-bit and 2-bit Characters (LeetCode #717) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The last character must always be a one-bit character if the pointer ends at the last index.

What is the time complexity of 1-bit and 2-bit Characters?

The optimal solution for 1-bit and 2-bit Characters runs in O(n) time and uses O(1) space. The complexity is O(n) because we only pass through the bits array once, making it efficient.

What is the brute force solution for 1-bit and 2-bit Characters?

The brute force approach for 1-bit and 2-bit Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will iterate through the bits array and try to decode each character based on whether it is a one-bit or two-bit character. This method is straightforward but inefficient as it checks each possible decoding. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve 1-bit and 2-bit Characters?

1-bit and 2-bit Characters uses the following concepts: Array. The recommended patterns to study are: Array, Greedy.

What are the interview tips for 1-bit and 2-bit Characters?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the minimum and maximum lengths of the input array. Explain your thought process while coding, as it helps the interviewer understand your approach.

What are common mistakes when solving 1-bit and 2-bit Characters?

Not accounting for the fact that the last character must be a one-bit character if the pointer ends at the last index. Confusing the conditions for moving the pointer, especially when encountering a '1'.

#717

1-bit and 2-bit Characters

Easy

ArrayArrayGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the array to determine if the last character is a one-bit character. By keeping track of the last valid character position, we can efficiently decide the outcome without unnecessary checks.

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Algorithm

4 steps

1Step 1: Initialize a pointer at the start of the bits array.
2Step 2: Iterate through the bits array until the second last element.
3Step 3: If the current bit is 1, skip the next bit (move the pointer by 2). If it's 0, move the pointer by 1.
4Step 4: At the end of the iteration, check if the pointer is at the last bit (0). If yes, return true; otherwise, return false.

solution.py5 lines

1def isOneBitCharacter(bits):
2    i = 0
3    while i < len(bits) - 1:
4        i += 2 if bits[i] == 1 else 1
5    return i == len(bits) - 1

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Complexity note: The complexity is O(n) because we only pass through the bits array once, making it efficient.

1The last character must always be a one-bit character if the pointer ends at the last index.
2The decoding is straightforward as we only need to check the current bit and decide how many bits to skip.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.