✅✅Easy C++ Solution || Simple to Understand
\tclass Solution {\n\tpublic:\n\t\tint countGoodRectangles(vector>& rectangles) {\n\n\t\t\tint n = rectangles.size();\n\t\t\tvector v;\n\n\t\t\tfor(int i=0 ; i<
\tclass Solution {\n\tpublic:\n\t\tint countGoodRectangles(vector<vector<int>>& rectangles) {\n\n\t\t\tint n = rectangles.size();\n\t\t\tvector<int> v;\n\n\t\t\tfor(int i=0 ; i<n ; i++){    \n\t\t\t\tv.push_back(min(rectangles[i][0], rectangles[i][1]));\n\t\t\t}\n\n\t\t\tint maxLen = *max_element(v.begin(), v.end());\n\n\t\t\treturn count(v.begin(), v.end(), maxLen);\n\t\t}\n\t};\nI hope that you\'ve found the solution useful.\nIn that case, please do upvote. Happy Coding :)
['C', 'C++']