Easy solution in hindi || faster than 100%
Intuition\nCommon square common rectangle bhi hota hai\nsquare find karne se aasan hai rectangle find karlo!!!\naur usse square nikalne ke liye choti vali side 
# Intuition\nCommon square common rectangle bhi hota hai\nsquare find karne se aasan hai rectangle find karlo!!!\naur usse square nikalne ke liye choti vali side ka square kar do simple\n\n# Approach\nToh common rectangle find karne ke liye bhi hamhe rectangle ka top aur bottom chahiye\ntoh bottom toh vohi common hoga naa jo upar(max) hoga(Tabhi toh common hoga)\ntop vohi common hoga jo dono rectangle mai neeche(min) hoga(graph dekho)\n\nab baaki comments code mai\n\n# Complexity\n- Time complexity:\nO(n^2)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\n    public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {\n        long ans=0;\n        for(int i=0;i<bottomLeft.length;i++){\n            //pehla rectangle\n            int b1[]=bottomLeft[i];\n            int t1[]=topRight[i];\n            for(int j=i+1;j<bottomLeft.length;j++){\n                //dusra rectangle   \n                int b2[]=bottomLeft[j];\n                int t2[]=topRight[j];\n                //bottom ke x and y ka max nikalo\n                int bx=Math.max(b1[0],b2[0]);\n                int by=Math.max(b1[1],b2[1]);\n                //top ke x and y ka min nikalo\n                int tx=Math.min(t1[0],t2[0]);\n                int ty=Math.min(t1[1],t2[1]);\n                //agar bottom top se bada hai matlab rectangle possible nahi\n                if(tx<bx||ty<by) continue;\n                else{\n                    //jo side choti hai rectangle ki usi size ka square ban sakta hai\n                    long side=Math.min(tx-bx,ty-by);\n                    ans=Math.max(ans,side*side);\n                }\n            }\n        }\n        return ans;\n    }\n}\n```\n\n\n\n**UPVOTE KAR DO PLEAASEEEEEEEEEEEEEEEEEEE!!!!!!!**
['Array', 'Math', 'Java']