How do you solve Tenth Line on LeetCode?

Tenth Line (LeetCode #195) 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: Reading files line by line is efficient for large files.

What is the brute force solution for Tenth Line?

The brute force approach for Tenth Line is Brute Force, with O(n) time complexity and O(1) space complexity. The simplest way to find the 10th line is to read through the file line by line and count until we reach the 10th line. If we reach the end of the file before the 10th line, we can handle that case separately. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Tenth Line?

Tenth Line uses the following concepts: Shell. The recommended patterns to study are: File I/O, Enumeration.

What are the interview tips for Tenth Line?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Practice reading files and handling exceptions in your code.

What are common mistakes when solving Tenth Line?

Not checking if the file exists before reading. Forgetting to handle the case where there are fewer than 10 lines.

#195

Tenth Line

Easy

ShellFile I/OEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of reading through the entire file, we can use a more efficient method by leveraging built-in functions that directly access the 10th line if it exists, reducing unnecessary reads.

⚙️

Algorithm

4 steps

1Step 1: Open the file for reading.
2Step 2: Use a buffer to read lines efficiently.
3Step 3: Read lines until the 10th line is reached or end of file.
4Step 4: If the 10th line is reached, print it; otherwise, handle the case where there are fewer than 10 lines.

solution.py8 lines

1# Full working Python code
2with open('file.txt', 'r') as file:
3    for i, line in enumerate(file):
4        if i == 9:
5            print(line.strip())
6            break
7    else:
8        print('File has less than 10 lines.')

ℹ

Complexity note: The time complexity remains O(n) as we still need to read lines, but we handle fewer lines in memory. Space complexity is O(1) as we only store a few variables.

1Reading files line by line is efficient for large files.
2Always handle edge cases, like fewer than 10 lines.

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