[C++] Two Solutions, One 2-liner
\nclass Solution {\npublic:\n    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {\n        return ((min(rec1[2], rec2[2]) - max(rec1[0], rec2[0])
```\nclass Solution {\npublic:\n    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {\n        return ((min(rec1[2], rec2[2]) - max(rec1[0], rec2[0]) > 0) &&\n                (min(rec1[3], rec2[3]) - max(rec1[1], rec2[1]) > 0));\n    }\n};\n```\n\n```\nclass Solution {\npublic:\n    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {\n        return ((rec2[0] < rec2[2]) && (rec2[1] < rec2[3]) &&\n                (rec1[0] < rec1[2]) && (rec1[1] < rec1[3]) && \n                (rec2[2] > rec1[0] && rec1[2] > rec2[0]) &&\n                (rec2[3] > rec1[1] && rec1[3] > rec2[1]));\n    }\n};\n```
['C']