CF 102697014 - Labyrinth
The labyrinth is represented by a rectangular grid of tiles. Some tiles contain traps, marked with x, while the remaining tiles are empty and marked with .. The task is simply to count how many trap tiles exist in the entire labyrinth and print that number.
Rating: -
Tags: -
Solve time: 1m 5s
Verified: yes
Solution
Problem Understanding
The labyrinth is represented by a rectangular grid of tiles. Some tiles contain traps, marked with x, while the remaining tiles are empty and marked with .. The task is simply to count how many trap tiles exist in the entire labyrinth and print that number. The input gives the height and width of the grid followed by every row of tiles. The output is the total number of positions containing a trap.
The constraints determine how much work is necessary. Since the input itself contains every tile, any solution must inspect the grid at least once. A linear scan over all n * m tiles is the natural target because it performs exactly the amount of work needed to read the data. Any approach that repeatedly searches the grid or compares every pair of cells would add unnecessary operations and could become too slow as the grid grows.
The main edge cases come from forgetting that every tile matters independently. A grid with no traps should produce zero, not a positive count caused by assuming at least one trap exists. For example:
1 3
...
The correct output is:
0
A careless implementation that initializes the answer to 1 would fail.
A single tile grid also needs to be handled correctly. For example:
1 1
x
The correct output is:
1
An implementation that assumes there are multiple rows or columns could access invalid positions.
A final common mistake is counting rows instead of cells. For example:
2 3
xxx
...
The correct output is:
3
The answer is the number of trap characters, not the number of rows containing at least one trap.
Approaches
The straightforward approach is to examine every tile and increase a counter whenever the current character is x. This method is already optimal because the input is a grid and every character has to be read. A brute force alternative could try to search for traps repeatedly, but that would only revisit information that has already been processed. If there are n * m cells, checking each cell once takes n * m operations, while any method that rescans the whole grid for every possible position could grow to (n * m)^2 operations.
The key observation is that traps have no relationship with each other. There is no movement, path finding, or dependency between cells. Each character contributes exactly one unit to the answer if and only if it is x. This allows the problem to be reduced to a single pass through the input.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O((n * m)^2) | O(1) | Too slow |
| Optimal | O(n * m) | O(1) | Accepted |
Algorithm Walkthrough
- Read the dimensions of the labyrinth and initialize a counter for the number of traps.
- Process each row of the grid. For every character in the row, add one to the counter when the character is
x. The decision is local because a tile either contains a trap or it does not. - Print the final counter value after every tile has been examined.
Why it works:
The invariant maintained during the scan is that the counter always equals the number of trap tiles among all cells processed so far. When a non trap tile is processed, the counter remains unchanged. When a trap tile is processed, the counter increases by exactly one, matching the contribution of that tile. After the final cell is processed, the counter therefore represents the number of traps in the entire labyrinth.
Python Solution
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
answer = 0
for _ in range(n):
row = input().strip()
answer += row.count('x')
print(answer)
The code reads each row exactly once and uses the built in string counting operation to count trap characters in that row. This follows the same single pass idea from the algorithm.
The width value is not used after input because the row itself already contains exactly the cells that must be inspected. The strip() call removes the newline character without changing the grid characters. There are no indexing operations, so boundary errors cannot occur.
Integer overflow is not a concern in Python because integers grow automatically. The maximum possible answer is the number of cells, which is already bounded by the input size.
Worked Examples
Consider this labyrinth:
4 5
x....
.x.x.
x..x.
x.xx.
The scan progresses as follows.
| Row | Trap count in row | Total traps |
|---|---|---|
x.... |
1 | 1 |
.x.x. |
2 | 3 |
x..x. |
2 | 5 |
x.xx. |
3 | 8 |
The final count is 8. This trace shows that every row contributes independently and the answer is simply the sum of trap characters.
A second example:
3 4
....
....
....
The scan is:
| Row | Trap count in row | Total traps |
|---|---|---|
.... |
0 | 0 |
.... |
0 | 0 |
.... |
0 | 0 |
The final answer remains zero. This demonstrates the empty labyrinth case where no special handling is needed.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n * m) | Every tile is inspected once while reading the grid. |
| Space | O(1) | Only the current row and the running counter are stored. |
The algorithm matches the amount of information provided by the input. Since every cell must be read, a linear scan is the best possible complexity and easily fits within the limits.
Test Cases
import sys
import io
def solve(inp: str) -> str:
old_stdin = sys.stdin
sys.stdin = io.StringIO(inp)
n, m = map(int, sys.stdin.readline().split())
ans = 0
for _ in range(n):
ans += sys.stdin.readline().strip().count('x')
sys.stdin = old_stdin
return str(ans) + "\n"
assert solve("""4 5
x....
.x.x.
x..x.
x.xx.
""") == "8\n", "sample 1"
assert solve("""3 4
....
....
....
""") == "0\n", "all empty"
assert solve("""1 1
x
""") == "1\n", "minimum grid with trap"
assert solve("""2 3
xxx
...
""") == "3\n", "one full row"
assert solve("""3 3
xxx
xxx
xxx
""") == "9\n", "all cells are traps"
| Test input | Expected output | What it validates |
|---|---|---|
4 5 sample grid |
8 |
Standard counting case |
| Empty grid | 0 |
Handles no traps correctly |
Single tile x |
1 |
Minimum size boundary |
| One trap row | 3 |
Counts cells instead of rows |
| Full grid of traps | 9 |
Maximum density of traps |
Edge Cases
For the empty labyrinth case:
1 3
...
the algorithm reads the only row, finds that it contains zero x characters, and leaves the counter unchanged. It prints 0, which matches the required result.
For the single tile case:
1 1
x
the algorithm processes one character, increases the counter once, and prints 1. No special condition is needed because the normal scan already handles the smallest possible grid.
For the case where several traps are on the same row:
2 3
xxx
...
the first row contributes three traps and the second row contributes none. The algorithm counts all three characters separately, avoiding the mistake of treating the row as a single object. The output is 3.