https://leetcode.com/problems/walking-robot-simulation/
https://leetcode.com/problems/walking-robot-simulation/discuss/155682/Logical-Thinking-with-Clear-Code
https://leetcode.com/problems/walking-robot-simulation/discuss/152364/Python-short-and-straightforward-solution-w-explanation-and-statement-is-wrong-in-the-question-!!!
A robot on an infinite grid starts at point (0, 0) and faces north. The robot can receive one of three possible types of commands:
-2: turn left 90 degrees-1: turn right 90 degrees1 <= x <= 9: move forwardxunits
Some of the grid squares are obstacles.
The
i-th obstacle is at grid point (obstacles[i][0], obstacles[i][1])
If the robot would try to move onto them, the robot stays on the previous grid square instead (but still continues following the rest of the route.)
Return the square of the maximum Euclidean distance that the robot will be from the origin.
Example 1:
Input: commands = [4,-1,3], obstacles = [] Output: 25 Explanation: robot will go to (3, 4)
Example 2:
Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]] Output: 65 Explanation: robot will be stuck at (1, 4) before turning left and going to (1, 8)
0 <= commands.length <= 100000 <= obstacles.length <= 10000-30000 <= obstacle[i][0] <= 30000-30000 <= obstacle[i][1] <= 30000- The answer is guaranteed to be less than
2 ^ 31.
- Time Complexity: , where are the lengths of
commandsandobstaclesrespectively. - Space Complexity: , the space used in storing the
obstacleSet.
The robot starts at point (0, 0) and faces north. Which edge of grid is to the north?
Since it will go to point (3, 4) with commands = [4,-1,3], obstacles = [], we know that the right edge is to the
Since it will go to point (3, 4) with commands = [4,-1,3], obstacles = [], we know that the right edge is to the
North. W
S -|- N
E
How do we represent absolute orientations given only relative turning directions(i.e., left or right)? We define
direction indicates the absolute orientation as below:North, direction = 0, directions[direction] = {0, 1}
East, direction = 1, directions[direction] = {1, 0}
South, direction = 2, directions[direction] = {0, -1}
West, direction = 3, directions[direction] = {-1, 0}
direction will increase by one when we turn right,
and will decrease by one when we turn left.
In this way, if the robot faces South, i.e., its direction is 2, when it moves forward by one step, x += 0, y += -1 for directions[2] = {0, -1}.
https://leetcode.com/problems/walking-robot-simulation/discuss/152322/Maximum!-This-is-crazy! public int robotSim(int[] commands, int[][] obstacles) {
Set<String> set = new HashSet<>();
for (int[] obs : obstacles) {
set.add(obs[0] + " " + obs[1]);
}
int[][] dirs = new int[][]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int d = 0, x = 0, y = 0, result = 0;
for (int c : commands) {
if (c == -1) {
d++;
if (d == 4) {
d = 0;
}
} else if (c == -2) {
d--;
if (d == -1) {
d = 3;
}
} else {
while (c-- > 0 && !set.contains((x + dirs[d][0]) + " " + (y + dirs[d][1]))) {
x += dirs[d][0];
y += dirs[d][1];
}
}
result = Math.max(result, x * x + y * y);
}
return result;
}- Firstly, robot doesn't face north as stated in the explanation. When i consider north, i imagine robot faces upwards but its initial direction is to right actually.
- Note: This is a similar question to Robot Room cleaner (No: 489).
- Code explanation:
- At first, we should make obstacles as set for O(1) check for obstacles and we simply move according to command if there is no obstacle.
- Critical part is the move array in my opinion. It is a simplified move to next i and j in array form for "right", "up", "left", "down" respectively.
- We also change d variable, which is direction variable as move index, when we get -2 or -1 command.
- If command is -2, we should turn left, which is to increment direction index.
- If command is -1, we should turn right, which is to decrement direction index.
- Else, we can walk through untill face an obstacle or not.
- And we update mx value in each move also.
public int robotSim(int[] commands, int[][] obstacles) {
int[] dx = new int[] { 0, 1, 0, -1 };
int[] dy = new int[] { 1, 0, -1, 0 };
int x = 0, y = 0, di = 0;
// Encode obstacles (x, y) as (x+30000) * (2^16) + (y+30000)
Set<Long> obstacleSet = new HashSet();
for (int[] obstacle : obstacles) {
long ox = (long) obstacle[0] + 30000;
long oy = (long) obstacle[1] + 30000;
obstacleSet.add((ox << 16) + oy);
}
int ans = 0;
for (int cmd : commands) {
if (cmd == -2) // left
di = (di + 3) % 4;
else if (cmd == -1) // right
di = (di + 1) % 4;
else {
for (int k = 0; k < cmd; ++k) {
int nx = x + dx[di];
int ny = y + dy[di];
long code = (((long) nx + 30000) << 16) + ((long) ny + 30000);
if (!obstacleSet.contains(code)) {
x = nx;
y = ny;
ans = Math.max(ans, x * x + y * y);
}
}
}
}
return ans;
}
