Question

You are given two arrays a and b of length n and m respectively
Consider the following (n + m) points on the coordinate plane:
|(~1,0), (a2,0), (a3,0)...... (an,0)]
|(61,1), (b2,1), (b3, 1)...... (bm, 1)]
Find the area of the largest rectangle that can be constructed from the given points​

Answer 1

Answer:

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Answer 2

Answer:

Check this out! I hope it helps you.

#include <iostream>

using namespace std;

int main()

{

int a[100],b[100],length,breadth1,breadth2,l1,k,j,l,i,area,y=0,z=1,t,m,n,max=0,s=1;

  cin>>t;

  while(s<=t)

  {

      cin>>m>>n;

      for(i=0;i<m;i++)

      {

          cin>>a[i];

      }

       for(j=0;j<n;j++)

      {

          cin>>b[j];

      }

   

       for(i=0;i<m;i++)

      {

         cout<<a[i];

      }

      cout<<endl;

       for(j=0;j<n;j++)

      {

         cout<<b[j];

      }

      cout<<endl;

      for(i=0,j=0;i<m,j<n;i++,j++)

      {

          for(k=i+1,l=j+1;k<m,l<n;k++,l++)

          {

          breadth1=a[k]-a[i];

          breadth2=b[l]-b[j];

           if(breadth1==breadth2)

           {

              length=z-y;

              area=length*breadth1;

              if(area>max)

              {

                 max=area;

              }

              break;

           }

         

          }

      }

      cout<<max;

      t++;

     

   }

 

}

Explanation:

Condition for being a rectangle is that the breadth of opposite sides is equal and the length of the opposite sides is equal.

So here the condition is 'a' is array representing (a1,0),(a2,0) and so on...

Taking y=0

and 'b' is another array representing (b1,1),(b2,1)....

taking z=1

so length will always be the difference of 1.

Hence, we only need to compare the rectangle's breadth

so take  two variables, compare their breadth by traversing array and find area.