[C++/Java/Python] Easy and Concise Solution
Your distance to target is abs(t[0]) + abs(t[1]).\nFor every ghost g, distance to target is abs(t[0] - g[0]) + abs(t[1] - g[1]).\nYou need to be closer to targe
Your distance to target is ```abs(t[0]) + abs(t[1])```.\nFor every ghost `g`, distance to target is `abs(t[0] - g[0]) + abs(t[1] - g[1])`.\nYou need to be closer to target than any ghost to escape.\n\nC++\n```\nbool escapeGhosts(vector<vector<int>>& ghosts, vector<int>& t) {\n        int d = abs(t[0]) + abs(t[1]);\n        for(auto g: ghosts)\n            if (d >= abs(t[0] - g[0]) + abs(t[1] - g[1]))\n                return false;\n        return true;\n    }\n```\nJava\n```\npublic boolean escapeGhosts(int[][] ghosts, int[] t) {\n        int d = Math.abs(t[0]) + Math.abs(t[1]);\n        for (int[] g: ghosts)\n            if (d >= Math.abs(t[0] - g[0]) + Math.abs(t[1] - g[1]))\n                return false;\n        return true;\n    }\n```\nPython:\n```\ndef escapeGhosts(self, ghosts, target):\n        x, y = target\n        d = abs(x) + abs(y)\n        return all(d < abs(x - i) + abs(y - j) for i, j in ghosts)\n```
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