Small observation using manhattan distance || Beats 100% || O(n)
Approach\n Describe your approach to solving the problem. \nFor who doesn\'t manhattan distance \n\nManhattan distance, represented by abs(x1-x2) + abs(y1-y2), 
# Approach\n<!-- Describe your approach to solving the problem. -->\nFor who doesn\'t manhattan distance \n\nManhattan distance, represented by abs(x1-x2) + abs(y1-y2), measures the sum of absolute differences in x and y coordinates between two points in a two-dimensional plane.\n\nFor reaching safely, the manhattan distance of the given point should be less than all the ghosts to target, or else they can reach to the final target before us.\n\nNow what if the distance of all ghosts to target is greater than ours? Because there is a restriction on ghosts and we can travel only one unit at a time, wherever they move, we can find some other way, and we can also reach our final destination without colliding with any of them.\n\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 {\npublic:\n    bool escapeGhosts(vector<vector<int>>& ghosts, vector<int>& target) {\n        int val = abs(target[0])+abs(target[1]) ; \n        for(auto it : ghosts){\n            int temp = abs(target[0]-it[0]) +abs(target[1]-it[1]) ; \n            if(temp <= val){\n                return false;\n            }\n        }\n        return true ;\n    }\n};\n```
['Math', 'C++']