LeetCode 194: Transpose File Solution

Master LeetCode problem 194 (Transpose File), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

194. Transpose File

Medium

Shell

Problem Explanation

Explanation

To transpose a file, we need to read the input file line by line and store each field in a 2D list where the rows represent the original columns and the columns represent the original rows. Then, we can iterate over this 2D list to generate the transposed content.

Time complexity: O(N*M) where N is the number of rows and M is the number of columns in the input file.
Space complexity: O(N*M) to store the transposed content.

Solution Code

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

public class TransposeFile {
    public static void main(String[] args) {
        List<List<String>> rows = new ArrayList<>();
        try (BufferedReader br = new BufferedReader(new FileReader("file.txt"))) {
            String line;
            while ((line = br.readLine()) != null) {
                String[] fields = line.split(" ");
                for (int i = 0; i < fields.length; i++) {
                    if (rows.size() <= i) {
                        rows.add(new ArrayList<>());
                    }
                    rows.get(i).add(fields[i]);
                }
            }
        } catch (IOException e) {
            e.printStackTrace();
        }

        try (FileWriter fw = new FileWriter("output.txt")) {
            for (List<String> row : rows) {
                fw.write(String.join(" ", row) + "\n");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

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