How do you solve Transpose File on LeetCode?

Transpose File (LeetCode #194) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transposing data is similar to flipping a matrix in mathematics.

What is the brute force solution for Transpose File?

The brute force approach for Transpose File is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves reading the entire file into memory, splitting the contents into rows and columns, and then manually constructing the transposed output. This is straightforward but inefficient for larger files. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Transpose File?

Transpose File uses the following concepts: Shell. The recommended patterns to study are: Array, Matrix Manipulation.

What are the interview tips for Transpose File?

Clarify assumptions about the input format. Discuss edge cases like empty files or single row/column. Explain your thought process clearly while coding.

What are common mistakes when solving Transpose File?

Not handling trailing spaces in output. Assuming variable number of columns in rows.

#194

Transpose File

Medium

ShellArrayMatrix Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution reads the file line by line and directly constructs the transposed output without storing all data in memory at once. This is efficient for large files.

⚙️

Algorithm

3 steps

1Step 1: Open the file and read the first line to determine the number of columns.
2Step 2: Initialize a list of strings to hold each transposed row.
3Step 3: For each subsequent line, split it into columns and append each column to the corresponding transposed row.

solution.py8 lines

1# Full working Python code
2with open('file.txt') as f:
3    lines = f.readlines()
4    transposed = [''] * len(lines[0].split())
5    for line in lines:
6        for i, word in enumerate(line.split()):
7            transposed[i] += word + ' '
8    print('\n'.join(row.strip() for row in transposed))

ℹ

Complexity note: The time complexity is O(n) because we read through the file once, and the space complexity is O(n) for storing the transposed rows.

1Transposing data is similar to flipping a matrix in mathematics.
2Reading files line by line is more memory efficient than loading the entire file at once.

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