✅ [C++] || Brute Force Simple Solution  || maxlength | maxbreadth
Now Comments will take over the audience to last line "return ans";\uD83D\uDE0C\n\nclass Solution {\npublic:\n    int maxArea(int h, int w, vector<int>& horizon
Now Comments will take over the audience to last line "return ans";\uD83D\uDE0C\n```\nclass Solution {\npublic:\n    int maxArea(int h, int w, vector<int>& horizontalCuts, vector<int>& verticalCuts) \n    {\n        int maxhori=0,maxverti=0,temp1=0,temp2=0;\n        sort(horizontalCuts.begin(),horizontalCuts.end()); // if not sorted then irregular maxlength is observed\n        sort(verticalCuts.begin(),verticalCuts.end());         // if not sorted then irregular maxbreadth is observed\n        for(int i=0;i<horizontalCuts.size();i++)\n        {\n            maxhori=max(maxhori,horizontalCuts[i]-temp1);       // we find max diff(maxlength) between intersecting points    hc[0]-0, hc[i]-hc[i-1], \n            temp1=horizontalCuts[i];\n        }\n        temp1=h;\n        maxhori=max(maxhori,temp1-horizontalCuts.back());         // h-hc[size-1]  may be it makes large\n        for(int i=0;i<verticalCuts.size();i++)\n        {\n            maxverti=max(maxverti,verticalCuts[i]-temp2);            //// we find max diff(maxbreadth) between intesecting points    vc[0]-0, vc[i]-vc[i-1], \n            temp2=verticalCuts[i];\n        }\n        temp2=w;\n        maxverti=max(maxverti,temp2-verticalCuts.back());     // w-vc[size-1]  may be it makes large \n        long long  int m=1000000007;\n        return ((maxhori%m)*(maxverti)%m);                      // to avoid Runtime error for large numbers\n    }\n};\n```\nIf you got it please UPVOTE!!
['Greedy', 'C', 'Sorting', 'C++']