Easiest Solution ! Beats 100% ! Simple Thought Process !
Intuition\n Describe your first thoughts on how to solve this problem. \nAs the ghosts move in the same time as the PAC-MAN, we only have to consider who would 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs the ghosts move in the same time as the PAC-MAN, we only have to consider who would reach destination first. Hence, simply calculate manhattan distance from (0,0) to target and also minTime for any ghost to reach destination. If time taken for PAC-MAN to reach destination is **strictly** **lesser** than minTime for ghost to reach, then it is possible to reach.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.Find manhattan distance from source to target.\n2.Find minimum time for any ghost to reach destination by calculating Manhattan Distance, and maintaining minimum distance for any ghost to destination.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(g) , where g is number of ghosts.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1) - Constant space as we are not using any other dynamic variables.\n\n# Code\n```java []\nclass Solution {\n    private int distance(int[] p1,int[] p2){\n        return Math.abs(p2[0] - p1[0]) + Math.abs(p2[1] - p1[1]);\n    }\n    public boolean escapeGhosts(int[][] ghosts, int[] target) {\n        //Calculate Manhattan Dist to target from all ghosts and from source\n        int distFromSrc = distance(new int[]{0,0} , target);\n\n        //Get least time for any ghost to reach dest\n        int minTime = Integer.MAX_VALUE;\n        for(int[] ghost : ghosts){\n            int dist = distance(ghost,target);\n            minTime = Math.min(minTime,dist);\n        }\n\n        return distFromSrc < minTime;\n        \n    }\n}\n```
['Array', 'Math', 'C++', 'Java']