# 874. Walking Robot Simulation

## Description

A robot on an infinite XY-plane starts at point (0, 0) facing north. The robot can receive a sequence of these three possible types of commands:

• -2: Turn left 90 degrees.
• -1: Turn right 90 degrees.
• 1 <= k <= 9: Move forward k units, one unit at a time.

Some of the grid squares are obstacles. The ith obstacle is at grid point obstacles[i] = (xi, yi). If the robot runs into an obstacle, then it will instead stay in its current location and move on to the next command.

Return the maximum Euclidean distance that the robot ever gets from the origin squared (i.e. if the distance is 5, return 25).

Note:

• North means +Y direction.
• East means +X direction.
• South means -Y direction.
• West means -X direction.
• There can be obstacle in [0,0].

Example 1:

Input: commands = [4,-1,3], obstacles = []
Output: 25
Explanation: The robot starts at (0, 0):
1. Move north 4 units to (0, 4).
2. Turn right.
3. Move east 3 units to (3, 4).
The furthest point the robot ever gets from the origin is (3, 4), which squared is 32 + 42 = 25 units away.


Example 2:

Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]]
Output: 65
Explanation: The robot starts at (0, 0):
1. Move north 4 units to (0, 4).
2. Turn right.
3. Move east 1 unit and get blocked by the obstacle at (2, 4), robot is at (1, 4).
4. Turn left.
5. Move north 4 units to (1, 8).
The furthest point the robot ever gets from the origin is (1, 8), which squared is 12 + 82 = 65 units away.


Example 3:

Input: commands = [6,-1,-1,6], obstacles = []
Output: 36
Explanation: The robot starts at (0, 0):
1. Move north 6 units to (0, 6).
2. Turn right.
3. Turn right.
4. Move south 6 units to (0, 0).
The furthest point the robot ever gets from the origin is (0, 6), which squared is 62 = 36 units away.


Constraints:

• 1 <= commands.length <= 104
• commands[i] is either -2, -1, or an integer in the range [1, 9].
• 0 <= obstacles.length <= 104
• -3 * 104 <= xi, yi <= 3 * 104
• The answer is guaranteed to be less than 231.

## Solutions

Solution 1: Hash table + simulation

We define a direction array $dirs=[0, 1, 0, -1, 0]$ of length $5$, where adjacent elements in the array represent a direction. That is, $(dirs[0], dirs[1])$ represents north, and $(dirs[1], dirs[2])$ represents east, and so on.

We use a hash table $s$ to store the coordinates of all obstacles, so that we can determine in $O(1)$ time whether the next step will encounter an obstacle.

We use two variables $x$ and $y$ to represent the current coordinates of the robot, initially $x = y = 0$. The variable $k$ represents the current direction of the robot, and the answer variable $ans$ represents the maximum Euclidean distance squared of the robot from the origin.

Next, we traverse each element $c$ in the array $commands$:

• If $c = -2$, it means that the robot turns left by $90$ degrees, that is, $k = (k + 3) \bmod 4$;
• If $c = -1$, it means that the robot turns right by $90$ degrees, that is, $k = (k + 1) \bmod 4$;
• Otherwise, it means that the robot moves forward by $c$ units of length. We combine the robot’s current direction $k$ with the direction array $dirs$, and we can get the increment of the robot on the $x$ axis and the $y$ axis. We add the increment of $c$ units of length to $x$ and $y$ respectively, and judge whether the new coordinates $(x, y)$ after each move are in the coordinates of obstacles. If not, update the answer $ans$, otherwise stop simulating and perform the simulation of the next instruction.

Finally return the answer $ans$.

Time complexity is $O(C \times n + m)$, space complexity is $O(m)$. Where $C$ represents the maximum number of steps that can be moved each time, and $n$ and $m$ respectively represent the lengths of arrays $commands$ and arrays $obstacles$.

• class Solution {
public int robotSim(int[] commands, int[][] obstacles) {
int[] dirs = {0, 1, 0, -1, 0};
Set<Integer> s = new HashSet<>(obstacles.length);
for (var e : obstacles) {
}
int ans = 0, k = 0;
int x = 0, y = 0;
for (int c : commands) {
if (c == -2) {
k = (k + 3) % 4;
} else if (c == -1) {
k = (k + 1) % 4;
} else {
while (c-- > 0) {
int nx = x + dirs[k], ny = y + dirs[k + 1];
if (s.contains(f(nx, ny))) {
break;
}
x = nx;
y = ny;
ans = Math.max(ans, x * x + y * y);
}
}
}
return ans;
}

private int f(int x, int y) {
return x * 60010 + y;
}
}

• class Solution {
public:
int robotSim(vector<int>& commands, vector<vector<int>>& obstacles) {
int dirs[5] = {0, 1, 0, -1, 0};
auto f = [](int x, int y) {
return x * 60010 + y;
};
unordered_set<int> s;
for (auto& e : obstacles) {
s.insert(f(e[0], e[1]));
}
int ans = 0, k = 0;
int x = 0, y = 0;
for (int c : commands) {
if (c == -2) {
k = (k + 3) % 4;
} else if (c == -1) {
k = (k + 1) % 4;
} else {
while (c--) {
int nx = x + dirs[k], ny = y + dirs[k + 1];
if (s.count(f(nx, ny))) {
break;
}
x = nx;
y = ny;
ans = max(ans, x * x + y * y);
}
}
}
return ans;
}
};

• class Solution:
def robotSim(self, commands: List[int], obstacles: List[List[int]]) -> int:
dirs = (0, 1, 0, -1, 0)
s = {(x, y) for x, y in obstacles}
ans = k = 0
x = y = 0
for c in commands:
if c == -2:
k = (k + 3) % 4
elif c == -1:
k = (k + 1) % 4
else:
for _ in range(c):
nx, ny = x + dirs[k], y + dirs[k + 1]
if (nx, ny) in s:
break
x, y = nx, ny
ans = max(ans, x * x + y * y)
return ans


• func robotSim(commands []int, obstacles [][]int) (ans int) {
dirs := [5]int{0, 1, 0, -1, 0}
type pair struct{ x, y int }
s := map[pair]bool{}
for _, e := range obstacles {
s[pair{e[0], e[1]}] = true
}
var x, y, k int
for _, c := range commands {
if c == -2 {
k = (k + 3) % 4
} else if c == -1 {
k = (k + 1) % 4
} else {
for ; c > 0 && !s[pair{x + dirs[k], y + dirs[k+1]}]; c-- {
x += dirs[k]
y += dirs[k+1]
ans = max(ans, x*x+y*y)
}
}
}
return
}

• function robotSim(commands: number[], obstacles: number[][]): number {
const dirs = [0, 1, 0, -1, 0];
const s: Set<number> = new Set();
const f = (x: number, y: number) => x * 60010 + y;
for (const [x, y] of obstacles) {
}
let [ans, x, y, k] = [0, 0, 0, 0];
for (let c of commands) {
if (c === -2) {
k = (k + 3) % 4;
} else if (c === -1) {
k = (k + 1) % 4;
} else {
while (c-- > 0) {
const [nx, ny] = [x + dirs[k], y + dirs[k + 1]];
if (s.has(f(nx, ny))) {
break;
}
[x, y] = [nx, ny];
ans = Math.max(ans, x * x + y * y);
}
}
}
return ans;
}