789. Escape The Ghosts
You are playing a simplified Pacman game. You start at the point
(0, 0)
, and your destination is(target[0], target[1])
. There are several ghosts on the map, the i-th ghost starts at(ghosts[i][0], ghosts[i][1])
.Each turn, you and all ghosts simultaneously *may* move in one of 4 cardinal directions: north, east, west, or south, going from the previous point to a new point 1 unit of distance away.
You escape if and only if you can reach the target before any ghost reaches you (for any given moves the ghosts may take.) If you reach any square (including the target) at the same time as a ghost, it doesn't count as an escape.
Return True if and only if it is possible to escape.
Example 1: Input: ghosts = [[1, 0], [0, 3]] target = [0, 1] Output: true Explanation: You can directly reach the destination (0, 1) at time 1, while the ghosts located at (1, 0) or (0, 3) have no way to catch up with you.Example 2: Input: ghosts = [[1, 0]] target = [2, 0] Output: false Explanation: You need to reach the destination (2, 0), but the ghost at (1, 0) lies between you and the destination.Example 3: Input: ghosts = [[2, 0]] target = [1, 0] Output: false Explanation: The ghost can reach the target at the same time as you.
Note:
- All points have coordinates with absolute value <=
10000
.- The number of ghosts will not exceed
100
.
Approach #1: Math. [Java]
class Solution { public boolean escapeGhosts(int[][] ghosts, int[] target) { int max = Math.abs(target[0]) + Math.abs(target[1]); for (int[] ghost : ghosts) { int dis = Math.abs(ghost[0] - target[0]) + Math.abs(ghost[1] - target[1]); if (dis <= max) return false; } return true; } }
Analysis:
The best tactic for any ghost is to reach the target before pacman and block the exit.
Note that we do not require that any ghost reaches pacman (which will never happen on an infinite grid for a single ghos and be much harder to determine for multiple ghost).
We only require that pacman can or cannot reach the target with optimal ghost strategy.
If any ghost has the same or lower distance to the target, then is can get there first and wait. Pacman cannot reach the target, although he would not necessarily be killed by a ghost.
If pacman is closer to the target than any ghost, he goes there along the most direct path.
Since we are working on a 2D grid, distances are measured as Manhattan distance.
Reference:
https://leetcode.com/problems/escape-the-ghosts/discuss/116511/Short-with-explanation-python