worst solution of mine
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{\npublic:\n    \n    long long inside (int a, int b, int c, int d, int p, int q, int r, int s)\n    {\n        long long X = 0, Y = 0;\n        \n        if ((a == p and c == r) or (a == p and c < r) or (p < a and c == r) or (p < a and c < r))\n        {\n            X = (c - a);\n        }\n        else if ((p == a and r < c) or (a < p and r == c) or (a < p and r < c))\n        {\n            X = (r - p);\n        } \n        else if (a < p and c < r and p < c)\n        {\n            X = (c - p);\n        }\n        else if (p < a and r < c and a < r)\n        {\n            X = (r - a);\n        }\n        \n        if ((b == q and d == s) or (b == q and d < s) or (q < b and d == s) or (q < b and d < s))\n        {\n            Y = (d - b);\n        }\n        else if ((b == q and s < d) or (b < q and d == s) or (b < q and s < d))\n        {\n            Y = (s - q);\n        }\n        else if ((b < q and d < s and q < d))\n        {\n            Y = (d - q);\n        }\n        else if (q < b and s < d and b < s)\n        {\n            Y = (s - b);\n        }\n        return min (X, Y) * min (X , Y);\n    }\n    \n    long long largestSquareArea (vector<vector<int>>& bL, vector<vector<int>>& tR) \n    {\n        int n = bL.size();\n        long long maxArea = 0;\n        \n        for (int i = 0; i < n; i++)\n        {\n            for (int j = i + 1; j < n; j++)\n            {\n                int a = bL[i][0], b = bL[i][1];\n                int c = tR[i][0], d = tR[i][1];\n                int p = bL[j][0], q = bL[j][1];\n                int r = tR[j][0], s = tR[j][1];\n\n                maxArea = max (inside (a, b, c, d, p, q, r, s), maxArea);\n            }\n        }\n        \n        return maxArea;\n    }\n};\n```
['C++']