CF 1468K - The Robot
We have a robot moving on an infinite 2D grid, starting at the origin (0, 0). It receives a sequence of movement commands: 'L' for left, 'R' for right, 'U' for up, and 'D' for down.
Rating: 1600
Tags: brute force, implementation
Solve time: 1m 32s
Verified: yes
Solution
Problem Understanding
We have a robot moving on an infinite 2D grid, starting at the origin (0, 0). It receives a sequence of movement commands: 'L' for left, 'R' for right, 'U' for up, and 'D' for down. Without any obstacles, executing all commands moves the robot to some final position, which is guaranteed to be not the origin. Our task is to place a single obstacle on a cell (not the origin) so that the robot, following the same commands, ends up back at (0, 0). If no such obstacle exists, we return (0, 0).
The input consists of multiple test cases. Each string of commands has length up to 5000, and the sum of lengths over all test cases is at most 5000. This constraint is crucial: it tells us we can afford an algorithm with time complexity roughly linear in the total number of commands. A brute-force solution that simulates the robot for every possible obstacle on the path could become quadratic, which would still be acceptable here because each string is small, but we can do better.
The edge cases that could trip a naive solution include very short sequences like "L" or "R", where the obstacle must be exactly one step away, and sequences where the robot moves in a loop or revisits the same cell, because multiple candidate obstacles could exist. Also, an obstacle that blocks a movement should only prevent that move, not teleport the robot or affect future movements beyond that step.
Approaches
A brute-force approach is straightforward: for every cell the robot visits along its path, simulate placing an obstacle there and then replay the entire command sequence. If the robot ends at the origin with that obstacle, we found a solution. This works because the robot path is short (at most 5000 commands), and the number of visited cells is at most the same. The worst-case operation count would be O(n²), where n is the length of the command string, because for each of up to n candidate cells, we simulate n moves. With the sum of lengths across all test cases ≤ 5000, this is fast enough.
The key insight to optimize is that we only need to check cells that the robot actually visits. Cells never visited cannot affect the robot. We can simulate the path once and store the sequence of positions. Then, for each visited position, we simulate the robot's path again, skipping the move that would step into that candidate obstacle. If after all moves the robot returns to the origin, we have a solution. This reduces unnecessary checks and is conceptually simple while remaining linear in practice for our input constraints.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(n²) | O(n) | Accepted due to small input constraints |
| Optimal | O(n²) | O(n) | Accepted, simpler to reason about and avoids impossible cells |
Algorithm Walkthrough
- Start by reading the number of test cases. For each test case, initialize the robot at (0, 0) and an empty list to record the path of all positions the robot visits.
- Simulate the robot’s movement command by command. For each move, update the current coordinates and append the new position to the path list.
- For each position in the path list, consider it as a candidate for placing the obstacle. Skip the origin because we cannot place the obstacle there.
- Simulate the robot’s movement again for the entire command string, but whenever a move would result in entering the candidate obstacle cell, the robot stays in place instead of moving.
- After completing the simulation for this candidate obstacle, check the final coordinates. If the robot ends up at the origin, print this obstacle as the solution and stop checking further candidates for this test case.
- If no candidate obstacle yields a return to the origin, output (0, 0) to indicate no solution exists.
Why it works: The algorithm maintains the invariant that the obstacle only affects one step at a time. Because we consider every cell the robot actually visits, we are guaranteed to find a cell that prevents the robot from reaching its original endpoint. If blocking a cell causes the robot to return to (0, 0), that satisfies the problem. We never consider irrelevant cells, so the simulation only examines plausible candidates.
Python Solution
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
commands = input().strip()
x, y = 0, 0
path = []
for c in commands:
if c == 'L':
x -= 1
elif c == 'R':
x += 1
elif c == 'U':
y += 1
elif c == 'D':
y -= 1
path.append((x, y))
found = False
for ox, oy in path:
if ox == 0 and oy == 0:
continue
rx, ry = 0, 0
for c in commands:
nx, ny = rx, ry
if c == 'L':
nx -= 1
elif c == 'R':
nx += 1
elif c == 'U':
ny += 1
elif c == 'D':
ny -= 1
if nx == ox and ny == oy:
continue
rx, ry = nx, ny
if rx == 0 and ry == 0:
print(ox, oy)
found = True
break
if not found:
print(0, 0)
The first loop computes the robot path. We store each visited cell in path to avoid considering impossible obstacle locations. In the second loop, we simulate again, adjusting the move only if it would hit the candidate obstacle. The final check ensures that the robot returns to the origin. We break immediately after finding a solution, as only one solution is required.
Worked Examples
Example 1: "L"
| Step | Command | x | y | Path |
|---|---|---|---|---|
| 0 | start | 0 | 0 | [] |
| 1 | L | -1 | 0 | [(-1,0)] |
Candidate obstacle (-1,0) prevents moving left. Simulating again, the robot stays at (0,0) and ends at origin. Output: -1 0.
Example 2: "RUUDL"
| Step | Command | x | y | Path |
|---|---|---|---|---|
| 0 | start | 0 | 0 | [] |
| 1 | R | 1 | 0 | [(1,0)] |
| 2 | U | 1 | 1 | [(1,0),(1,1)] |
| 3 | U | 1 | 2 | [(1,0),(1,1),(1,2)] |
| 4 | D | 1 | 1 | [(1,0),(1,1),(1,2),(1,1)] |
| 5 | L | 0 | 1 | [(1,0),(1,1),(1,2),(1,1),(0,1)] |
Candidate obstacle (1,2) prevents stepping into (1,2). Simulating again, the final coordinates are (0,0). Output: 1 2.
These traces confirm that the algorithm correctly identifies an obstacle that forces the robot back to the origin.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n²) | For each visited cell (up to n), we simulate the entire command sequence (length n). Worst case n² per test case. Sum of all commands ≤ 5000, so total ≤ 25,000,000 operations. |
| Space | O(n) | We store the path of all visited cells, which is at most the length of the command string. |
The algorithm fits comfortably within the 2-second time limit and 512 MB memory limit.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
output = io.StringIO()
sys.stdout = output
# paste solution here
t = int(input())
for _ in range(t):
commands = input().strip()
x, y = 0, 0
path = []
for c in commands:
if c == 'L':
x -= 1
elif c == 'R':
x += 1
elif c == 'U':
y += 1
elif c == 'D':
y -= 1
path.append((x, y))
found = False
for ox, oy in path:
if ox == 0 and oy == 0:
continue
rx, ry = 0, 0
for c in commands:
nx, ny = rx, ry
if c == 'L':
nx -= 1
elif c == 'R':
nx += 1
elif c == 'U':
ny += 1
elif c == 'D':
ny -= 1
if nx == ox and ny == oy:
continue
rx, ry = nx, ny
if rx == 0 and ry == 0:
print(ox, oy)
found = True
break
if not found:
print(0, 0)
return output.getvalue().strip()