C++ simple 1 liner
To simplify the process, instead of thinking about how the rectangles will intersect, think about how they will not, and negate that result. There are only two 
To simplify the process, instead of thinking about how the rectangles will intersect, think about how they will *not*, and negate that result. There are only two possible scenarios for when rectangles do not intersect, and that is if one rectangle is completely on top of the other, or one rectangle is completely to the left of the other. Because both `rec1` and `rec2` could be "that" rectangle, there will be a total of 4 conditions. \n\nRun-time is `O(1)`. \n\n```\nclass Solution {\npublic:\n    static inline constexpr bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {\n        return (rec1[1] >= rec2[3] || rec1[3] <= rec2[1]) ? false : !(rec1[0] >= rec2[2] || rec1[2] <= rec2[0]);\n    }\n};\n```
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