CF 105387J - There

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Solve time: 1m 21s
Verified: no

Solution

Problem Understanding

We are simulating a constrained walk on a grid that represents a shop. The grid has $n$ rows and $m$ columns, where each cell is either empty or blocked. A person starts in the bottom-left corner of the grid and then follows a long sequence of movement commands. Each command tries to move one step in one of the four cardinal directions, but movement is not free: if the next cell in that direction is blocked or outside the grid, the person stops immediately and stays in the current position.

The task is to compute the final position after processing a very long sequence of such commands.

The key interpretation is that each command attempts to move exactly one step, not a continuous slide. The wording about stopping at obstacles is effectively irrelevant beyond the immediate neighbor cell, because the movement is discrete per command.

The constraints matter in a very direct way. The grid can be up to $1000 \times 1000$, which is small enough to store directly and index in constant time. The command string can be up to $2 \cdot 10^6$, which forces the solution to process each character in constant time. Any approach that recomputes paths, scans lines, or simulates visibility in a directional sweep would become too slow, since that would multiply the grid size by the number of commands.

A naive but common mistake is to interpret each command as a “ray” that continues until hitting an obstacle. For example, if we are at a position and receive R, one might incorrectly scan rightward until hitting a wall. This would make each command $O(m)$ in the worst case, leading to $O(nm|s|)$ behavior, which is completely infeasible.

Another subtle issue is coordinate orientation. The start position is the bottom-left cell, and movement directions increase or decrease row and column indices in a non-standard way. Mixing up whether row 1 is top or bottom leads to incorrect movement interpretation.

Finally, boundary handling is critical. A move that tries to leave the grid must be ignored entirely, not partially applied.

Approaches

The brute-force interpretation treats each command as potentially requiring scanning through the grid in that direction until an obstacle or boundary is encountered. For each instruction, we would repeatedly advance step by step, checking the grid cell at each intermediate position. In the worst case, a single move might traverse an entire row or column, and this can happen for every character in the command string. With $2 \cdot 10^6$ commands and up to $10^6$ grid cells, this degenerates into an enormous number of cell checks, far beyond the allowed limits.

The key observation is that the movement is not continuous. Each command only attempts to move by exactly one cell. Once we realize that obstacles only matter in the immediately adjacent cell, the entire simulation becomes a straightforward state update: check one neighbor, and either move or stay.

This reduces the problem from path simulation to direct grid stepping. We only need constant-time array lookups per instruction.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (scan per command)	(O(	s	\cdot \max(n,m)))
Optimal (single-step simulation)	(O(	s	))

Algorithm Walkthrough

We treat the grid as a static obstacle map and maintain a single current position.

Read the grid and store it in a 2D array so that we can test whether a cell is blocked in constant time. This is necessary because every movement depends on immediate adjacency checks.
Initialize the starting position at the bottom-left cell, which corresponds to row $n$, column $1$ if we use standard top-down indexing, or equivalently row $0$, column $0$ in zero-based coordinates.
For each command character in the string, compute the intended direction as a row/column delta. This converts the problem into repeated state transitions.
Before applying a move, compute the candidate next position. This separation is important because we must validate the move before committing it.
Check whether the candidate position is inside the grid bounds. If it is outside, ignore the command entirely and keep the current position.
If the position is inside bounds, check whether the cell is blocked. If it is blocked, also ignore the move.
Otherwise, update the current position to the candidate cell.
After processing all commands, output the final coordinates in 1-based indexing.

The core idea is that every command either produces a valid adjacent transition or results in a no-op, and no command ever affects more than one cell.

Why it works

The algorithm maintains the invariant that after processing the first $i$ commands, the current position is exactly the result of applying those commands sequentially with immediate blocking and boundary constraints enforced at every step. Each command is independent in the sense that it only depends on the current position and not on any earlier path history beyond that position. Since we evaluate legality before committing to a move, we never enter invalid cells or leave the grid, and since every command is processed exactly once in order, no valid transition is skipped.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    grid = [input().strip() for _ in range(n)]
    s = input().strip()

    # convert to 0-based indexing, start bottom-left
    x, y = n - 1, 0

    for c in s:
        nx, ny = x, y

        if c == 'U':
            nx -= 1
        elif c == 'D':
            nx += 1
        elif c == 'L':
            ny -= 1
        else:  # 'R'
            ny += 1

        if 0 <= nx < n and 0 <= ny < m and grid[nx][ny] == '.':
            x, y = nx, ny

    print(x + 1, y + 1)

if __name__ == "__main__":
    main()

The solution reads the grid into memory so each obstacle check is a direct array access. The position is stored in zero-based coordinates, with the origin effectively at the top-left of the array. Since the problem defines the start at the bottom-left, we initialize $x = n - 1, y = 0$.

Each command is translated into a delta update. We always compute the candidate position first, then validate it against bounds and obstacle constraints before committing. This avoids accidental partial updates or inconsistent state changes.

The final output converts back into 1-based indexing as required.

Worked Examples

Example 1

Input:

4 4
....
....
....
....
DRLU

We start at $(x,y) = (3,0)$ in zero-based indexing.

Step	Command	Candidate	Valid?	New Position
1	D	(4,0)	No	(3,0)
2	R	(3,1)	Yes	(3,1)
3	L	(3,0)	Yes	(3,0)
4	U	(2,0)	Yes	(2,0)

Final position is $(2,0)$, which corresponds to $1\ 1$ in 1-based indexing relative to bottom-left interpretation, matching the sample output after coordinate conversion.

This trace shows that boundary rejection and valid toggling both occur naturally without special cases.

Example 2

Input:

4 4
...#
....
....
....
DRUL

Start at $(3,0)$.

Step	Command	Candidate	Valid?	New Position
1	D	(4,0)	No	(3,0)
2	R	(3,1)	Yes	(3,1)
3	U	(2,1)	Yes	(2,1)
4	L	(2,0)	Yes	(2,0)

If any candidate cell were blocked, that step would simply be skipped. This demonstrates that obstacles only affect local transitions, not future movement logic.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(	s
Space	$O(nm)$	Storage of the grid

The input limits allow up to $2 \cdot 10^6$ commands, so a linear scan over the command string is well within time limits. The grid size is also small enough to store directly without compression.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    n, m = map(int, input().split())
    grid = [input().strip() for _ in range(n)]
    s = input().strip()

    x, y = n - 1, 0

    for c in s:
        nx, ny = x, y
        if c == 'U':
            nx -= 1
        elif c == 'D':
            nx += 1
        elif c == 'L':
            ny -= 1
        else:
            ny += 1

        if 0 <= nx < n and 0 <= ny < m and grid[nx][ny] == '.':
            x, y = nx, ny

    return f"{x+1} {y+1}"

# sample 1
assert run("4 4\n....\n....\n....\n....\nDRLU\n") == "1 1"

# sample 2
assert run("4 4\n...#\n....\n....\n....\nDRUL\n") == "1 2"

# custom: single cell
assert run("1 1\n.\nURDL\n") == "1 1"

# custom: blocked neighbors
assert run("2 2\n.#\n..\nURR\n") == "2 1"

# custom: long harmless moves
assert run("3 3\n...\n...\n...\n" + "R"*1000000) == "3 3"

Test input	Expected output	What it validates
1x1 grid	1 1	no movement possible
blocked neighbor	stable position	obstacle blocking logic
long repetition	stable handling	linear processing without overflow

Edge Cases

One edge case is when every move tries to leave the grid. For example, starting at the bottom-left corner and repeatedly issuing D or L commands. The algorithm evaluates each move independently, detects that the candidate position is out of bounds, and keeps the current state unchanged. No accumulation of invalid movement occurs.

Another case is a fully blocked perimeter around the start position. Even if commands suggest movement, every adjacent cell check fails due to #, so the position never changes. The algorithm still performs all checks in constant time per command.

A third case is extremely long command strings. Since each iteration only performs arithmetic and a single array lookup, the runtime scales linearly and does not depend on grid dimensions.