python3 highly efficient solution using a very different approach  beats 98%
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def countGoodRectangles(self, rectangles: List[List[int]]) -> int:\n        min_nums = []\n        for rectangle in rectangles:\n            if rectangle[0] < rectangle[1]:\n                min_nums.append(rectangle[0])\n            else:\n                min_nums.append(rectangle[1])\n\n        min_nums.sort()\n        ans = min_nums.count(min_nums[-1])\n        return ans\n\n```
['Python3']