Algorithm
Problem Name: 789. Escape The Ghosts
You are playing a simplified PAC-MAN game on an infinite 2-D grid. You start at the point [0, 0], and you are given a destination point target = [xtarget, ytarget] that you are trying to get to. There are several ghosts on the map with their starting positions given as a 2D array ghosts, where ghosts[i] = [xi, yi] represents the starting position of the ith ghost. All inputs are integral coordinates.
Each turn, you and all the ghosts may independently choose to either move 1 unit in any of the four cardinal directions: north, east, south, or west, or stay still. All actions happen simultaneously.
You escape if and only if you can reach the target before any ghost reaches you. If you reach any square (including the target) at the same time as a ghost, it does not count as an escape.
Return true if it is possible to escape regardless of how the ghosts move, otherwise return false.
Example 1:
Input: ghosts = [[1,0],[0,3]], target = [0,1] Output: true Explanation: You can reach the destination (0, 1) after 1 turn, while the ghosts located at (1, 0) and (0, 3) cannot 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.
Constraints:
1 <= ghosts.length <= 100ghosts[i].length == 2-104 <= xi, yi <= 104- There can be multiple ghosts in the same location.
target.length == 2-104 <= xtarget, ytarget <= 104
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean escapeGhosts(int[][] ghosts, int[] target) {
int[] currPos = {0, 0};
int selfDist = getDist(currPos, target);
for (int[] ghost : ghosts) {
int ghostDist = getDist(ghost, target);
if (ghostDist < = selfDist) {
return false;
}
}
return true;
}
private int getDist(int[] p1, int[] p2) {
return Math.abs(p1[0] - p2[0]) + Math.abs(p1[1] - p2[1]);
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
var escapeGhosts = function(ghosts, target) {
let res = true
const { abs } = Math, steps = abs(target[0]) + abs(target[1])
const [tx, ty] = target
for(const [x, y] of ghosts) {
if(abs(tx - x) + abs(ty - y) <= steps> return false
}
return res
};
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def escapeGhosts(self, ghosts, target):
d = abs(target[0]) + abs(target[1])
for ghost in ghosts:
if abs(ghost[0] - target[0]) + abs(ghost[1] - target[1]) <= d: return False
return True
Input
Output