Simple JAVA Solution. 100%
Approach\n Describe your approach to solving the problem. \n    First it looks like a BFS graph problem. \n    But After drawing on paper, Comes out an easy pro
# Approach\n<!-- Describe your approach to solving the problem. -->\n    First it looks like a BFS graph problem. \n    But After drawing on paper, Comes out an easy problem.\n    Just try to go to target before the ghosts go to the target.\n    (I know they are not following the target but the PC man).\n\n    Give a thought like : If the ghost is at target position, then we always lose.\n    If it\'s in the way of PC Man or at a position closr to target than the PC man.. then it may catch It. or atleast go and sit at the target position. We Surely lose.\n    If it\'s at a far position, Maybe just a single point behind, then it will never catch the PC man, We win.\n\n    Check the abs distance of all ghosts from the target and compare with PC Man. \n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public boolean escapeGhosts(int[][] ghosts, int[] target) {\n        int pcManDis = getDis(target, new int[]{0,0});\n\n        for(int[] i: ghosts){\n            int ghostDis = getDis(target, i);\n\n            if(ghostDis <= pcManDis)\n                return false;\n        }\n        return true;\n       \n    }\n     private int getDis(int[] start, int[] end){\n\n            return (Math.abs(start[0] - end[0]) + Math.abs(start[1] - end[1]));\n\n        }\n}\n```
['Java']