Easy Iterative Solution (100% faster)
Intuition\nWe will iterate all possible pairs and find out the maximum square area that can be fit within the rectangles.\n\n# Approach\n\nIn the given code we 
# Intuition\nWe will iterate all possible pairs and find out the maximum square area that can be fit within the rectangles.\n\n# Approach\n\nIn the given code we iterate throught all possible pairs and find maxx, minx, maxy, and miny represent the coordinates of the rectangle formed by the overlap of two rectangles. Here\'s their significance:\n\nmaxx: Represents the maximum x-coordinate of the overlapping region. It is calculated as the minimum of the x-coordinate of the top-right corner of each rectangle. This ensures that maxx defines the rightmost boundary of the overlapping area.\n\nminx: Represents the minimum x-coordinate of the overlapping region. It is calculated as the maximum of the x-coordinate of the bottom-left corner of each rectangle. This ensures that minx defines the leftmost boundary of the overlapping area.\n\nmaxy: Represents the maximum y-coordinate of the overlapping region. It is calculated as the minimum of the y-coordinate of the top-right corner of each rectangle. This ensures that maxy defines the topmost boundary of the overlapping area.\n\nminy: Represents the minimum y-coordinate of the overlapping region. It is calculated as the maximum of the y-coordinate of the bottom-left corner of each rectangle. This ensures that miny defines the bottommost boundary of the overlapping area.\n\n\nAfter that we can make a check that minx and miny are less than maxx and maxy respectively and find out max area of square.\n\n\n\n# Complexity\n- Time complexity:\nO(N*N)\n- Space complexity:\nO(1)\n# Code\n```\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        long long res=0;\n\n        int n=bottomLeft.size();\n        for(int i=0;i<n;i++){\n            for(int j=i+1;j<n;j++){\n                int minx=max(bottomLeft[i][0],bottomLeft[j][0]);\n                int miny=max(bottomLeft[i][1],bottomLeft[j][1]);\n                int maxx=min(topRight[i][0],topRight[j][0]);\n                int maxy=min(topRight[i][1],topRight[j][1]);\n\n                if(minx<maxx && miny<maxy){\n                    long long side=min(maxx-minx,maxy-miny);\n                    res=max(res,side*side);\n                }\n            }\n        }\n        return res;\n    }\n};\n```
['C++']