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

The optimal solution has a time complexity of O(n).

What data structures are used to solve 1-bit and 2-bit Characters?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving 1-bit and 2-bit Characters?

Insight 1: The core idea is to simulate the decoding process from the beginning or the end of the array. Insight 2: We can iterate through the array, incrementing the index by 1 if we encounter '0' (one-bit character) and by 2 if we encounter '1' (two-bit character).

Which companies ask LeetCode 717. 1-bit and 2-bit Characters in interviews?

This problem is frequently asked at IXL, Quora.

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1-bit and 2-bit Characters - LeetCode 717 Solution

1-bit and 2-bit Characters - Complete Solution Guide

1-bit and 2-bit Characters is LeetCode problem 717, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 o

Detailed Explanation

The problem asks us to determine if the last character in a given binary array `bits` is a one-bit character (represented by '0'). The array `bits` contains characters that can be either one-bit ('0') or two-bit ('10' or '11'). The array is guaranteed to end with '0'. The task is to return `true` if the very last character is a one-bit character and `false` otherwise. For instance, if `bits = [1, 0, 0]`, it can be decoded as a two-bit character ('10') followed by a one-bit character ('0'). Thus, the answer should be `true`. If `bits = [1, 1, 1, 0]`, it's decoded as two two-bit characters ('11' and '10'), so the last character is part of a two-bit character, meaning the answer is `false`.

Solution Approach

The provided Python, C, and Java solutions implement two different but correct approaches. The Python/C solution iterates from the beginning of the array, simulating the decoding process. The Java and C++ solutions iterate backward from the second to last element to count the number of consecutive 1s.

Step-by-Step Algorithm

Step 1 (Python/C): Initialize an index `i` to 0.
Step 2 (Python/C): Iterate through the array while `i` is less than `len(bits) - 1`.
Step 3 (Python/C): If `bits[i]` is 1 (a two-bit character), increment `i` by 2.
Step 4 (Python/C): If `bits[i]` is 0 (a one-bit character), increment `i` by 1.
Step 5 (Python/C): After the loop, return `true` if `i` is equal to `len(bits) - 1`, indicating that the last character is a one-bit character. Otherwise, return `false`.
Step 1 (Java/C++): Start from the second to last element (index = `bits.length - 2`).
Step 2 (Java/C++): Iterate backwards while the current element is 1 and the index is greater than or equal to 0.
Step 3 (Java/C++): Count how many consecutive 1s are before the last 0. The last bit is a one-bit character if the count is even.

Key Insights

Insight 1: The core idea is to simulate the decoding process from the beginning or the end of the array.
Insight 2: We can iterate through the array, incrementing the index by 1 if we encounter '0' (one-bit character) and by 2 if we encounter '1' (two-bit character).
Insight 3: Alternatively, we can iterate from the second to last element backwards. If we encounter multiple consecutive 1s before the last 0, an odd number means the last zero is part of 2-bit character and hence can not be a one-bit character. Even number means it is a one-bit character.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: IXL, Quora.

1-bit and 2-bit Characters | LeetCode 717

Easy

Array

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int i = bits.length - 2;
        while (i >= 0 && bits[i] == 1) {
            i--;
        }
        return (bits.length - 1 - i) % 2 == 0;
    }
}

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Problem Info

Description

Problem 717: 1-bit and 2-bit Characters - Detailed explanation and solution approaches.

Topics

Array

Companies

IXLQuora

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:717

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1 (Python/C): Initialize an index `i` to 0.
2
Step 2 (Python/C): Iterate through the array while `i` is less than `len(bits) - 1`.
3
Step 3 (Python/C): If `bits[i]` is 1 (a two-bit character), increment `i` by 2.
4
Step 4 (Python/C): If `bits[i]` is 0 (a one-bit character), increment `i` by 1.
5
Step 5 (Python/C): After the loop, return `true` if `i` is equal to `len(bits) - 1`, indicating that the last character is a one-bit character. Otherwise, return `false`.
6
Step 1 (Java/C++): Start from the second to last element (index = `bits.length - 2`).
7
Step 2 (Java/C++): Iterate backwards while the current element is 1 and the index is greater than or equal to 0.
8
Step 3 (Java/C++): Count how many consecutive 1s are before the last 0. The last bit is a one-bit character if the count is even.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the algorithm iterates through the array at most once. The space complexity is O(1) because the algorithm only uses a few integer variables, which take up a constant amount of space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A recursive approach could also be used to solve the problem, where each recursive call checks if the current bit is a one-bit or two-bit character and then makes a recursive call for the remaining bits.

Time: O(n) in the worst case.Space: O(n) due to recursion depth in the worst case, which may lead to stack overflow issues for very large inputs.

Trade-offs: The recursive approach is less efficient than the iterative approach due to the overhead of function calls and the potential for stack overflow errors. The iterative approach is generally preferred.

✨

Key Insights

1Insight 1: The core idea is to simulate the decoding process from the beginning or the end of the array.
2Insight 2: We can iterate through the array, incrementing the index by 1 if we encounter '0' (one-bit character) and by 2 if we encounter '1' (two-bit character).
3Insight 3: Alternatively, we can iterate from the second to last element backwards. If we encounter multiple consecutive 1s before the last 0, an odd number means the last zero is part of 2-bit character and hence can not be a one-bit character. Even number means it is a one-bit character.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `bits = [0]`. The algorithm correctly identifies the last character as a one-bit character.
Edge case 2: `bits = [1, 0]`. The algorithm correctly identifies the last character as a one-bit character because '10' represents a two-bit character.
Edge case 3: `bits = [1, 1, 0]`. The algorithm correctly identifies the last character as a one-bit character because '11' represents a two-bit character. After consuming the first '1', the index moves to 1. bits[1] is '1' which is another two-bit, so we can only have two two-bits or one-bit at the end

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where there are consecutive 1s at the beginning of the array. The solution needs to correctly handle these consecutive 1s to determine if the last character is a one-bit character.

✗Common mistake 2: Assuming that a sequence like [1, 1, 1, 0] will always result in `false`. We must correctly account for all the consecutive 1s.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This problem can be related to decoding variable-length codes, like Huffman coding, where symbols have different bit representations.

🏦Real-world scenario 2: In network protocols, similar patterns can be used to identify message boundaries or different types of data packets. This problem offers a simplified version of the same concept.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "1-bit and 2-bit Characters"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?