Maximal Rectangle in C++

C++Server Side ProgrammingProgramming

Suppose we have a 2D binary matrix where 0s and 1 values are present. We have to find the largest rectangle containing only 1s and return its area.

To solve this, we will follow these steps−

  • Define a function called getAns, this will take array a

  • create stack st, i := 0, ans := 0

  • while i < size of a, then

    • if stack is empty or a[i] >= top of stack, then insert i into st, increase i by 1

    • otherwise −

      • height := a[top of stack], delete from stack

      • width := i when stack is empty, otherwise i – top of st – 1

      • area := height * width

      • ans := max of ans and area

  • while st is not empty

    • height := a[top of st], delete from stack

    • width := size of a when st is empty, otherwise size of a – top of st – 1

    • area := height * width

    • ans := max of ans and area

  • return ans

  • From the main method do the following −

  • ans := 0, n := size of x

  • if n zero, then return 0

  • m := size of x[0]

  • create one array height of size m

  • for i in range 0 to n – 1

    • for j in range 0 to m – 1

      • if x[i, j] = 1, then increase height[j] by 1, otherwise height[j] := 0

    • ans := max of ans and getAns(height)

  • return ans

Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int getAns(vector <int> a){
      stack <int> st;
      int i = 0;
      int ans = 0;
      while(i<a.size()){
         if(st.empty()||a[i]>=a[st.top()]){
            st.push(i);
            i++;
         } else{
            int height = a[st.top()];
            st.pop();
            int width = st.empty()?i:i-st.top()-1;
            int area = height * width;
            ans = max(ans,area);
         }
      }
      while(!st.empty()){
         int height = a[st.top()];
         st.pop();
         int width = st.empty()?a.size():a.size() - st.top()-1;
         int area = height * width;
         ans = max(ans,area);
      }
      return ans;
   }
   int maximalRectangle(vector<vector<char>>& x) {
      int ans = 0;
      int n = x.size();
      if(!n)return 0;
      int m = x[0].size();
      vector <int> height(m);;
      for(int i =0;i<n;i++){
         for(int j =0;j<m;j++){
            if(x[i][j] == '1')height[j]++;
            else height[j] = 0;
         }
         ans = max(ans, getAns(height));
      }
      return ans;
   }
};
main(){
   vector<vector<char>> v = {
      {'1','0','1','0','0'},
      {'1','0','1','1','1'},
      {'1','1','1','1','1'},
      {'1','0','0','1','0'}
   };
   Solution ob;
   cout << (ob.maximalRectangle(v));
}

Input

{{'1','0','1','0','0'},
{'1','0','1','1','1'},
{'1','1','1','1','1'},
{'1','0','0','1','0'}
}

Output

6
raja
Published on 26-May-2020 12:34:46
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