Overview

We check whether each row contains T, which represents a treasure. If it contains T, we assign o; otherwise, we assign x. We then concatenate these characters from the top row to the bottom row and output the resulting string.

Analysis

Whether a row contains a treasure can be determined solely by inspecting the string representing that row. Furthermore, the number and positions of the treasures are not required; only whether there is at least \(1\) T matters.

In Python, you can check whether a string s contains T using 'T' in s.

For example, suppose the rows are as follows:

..T.
....
T..T

• Row \(1\) contains T, so it corresponds to o.
• Row \(2\) does not contain T, so it corresponds to x.
• Row \(3\) contains T, so it corresponds to o.

Therefore, the output is oxo.

In this problem, even if we inspect all cells naively, there are at most \(10 \times 10 = 100\) cells, so it will not cause a TLE. No special optimizations are needed. Note, however, that you must output exactly \(1\) character per row, rather than outputting o for each treasure.

Algorithm

1. Split the input by whitespace or newlines to obtain the number of rows \(H\) and the string for each row.
2. From the top row to the bottom row, evaluate each string s:
   • If 'T' in s is true, choose o.
   • If false, choose x.
3. Concatenate the \(H\) chosen characters using join and print the result.

Since we select a character corresponding to the presence or absence of T in each row and arrange them in the same order as the input, we obtain the desired string.

Complexity

• Time Complexity: \(O(HW)\)
  Checking whether each row contains T takes at most \(O(W)\), and this is performed for \(H\) rows.
• Space Complexity: \(O(HW)\)
  The provided code reads and stores the entire input. An additional \(O(H)\) space is used for the output string.

Implementation Details

• sys.stdin.read().split() splits the input by whitespace and newlines. data[0] is \(H\), data[1] is \(W\), and data[2] onward are the strings for each row.

• data[2:2 + H] retrieves the \(H\) rows of the map in input order.

• Since \(W\) is given as the length of each string, there is no need to store it in a variable for this check.

• By using ''.join(...), you can construct a string of length \(H\) without spaces between o and x.

  Source Code

import sys

def main():
    data = sys.stdin.read().split()
    H = int(data[0])
    # W is given but not needed beyond parsing
    rows = data[2:2 + H]
    print(''.join('o' if 'T' in s else 'x' for s in rows))

main()

This editorial was generated by gpt-6-astra-medium.