shares a nice solution with explanation using DP. You will be clear of the algorithm after running it on its suggested example:

matrix = [[0, 0, 0, 1, 0, 0, 0],[0, 0, 1, 1, 1, 0, 0],[0, 1, 1, 1, 1, 1, 0]];

The code is rewritten as follows.

1 class Solution { 2 public: 3     int maximalRectangle(vector
     
      
       >& matrix) { 4         if (matrix.empty()) return 0; 5         const int m = matrix.size(), n = matrix[0].size(); 6         int *left = new int[n](), *right = new int[n](), *height = new int[n](); 7         fill_n(right, n, n); 8         int area = 0; 9         for (int i = 0; i < m; i++) {10             int l = 0, r = n;11             for (int j = 0; j < n; j++)12                 height[j] += matrix[i][j] == '1' ? 1 : -height[j];13             for (int j = 0; j < n; j++) {14                 if (matrix[i][j] == '1') left[j] = max(left[j], l);15                 else left[j] = 0, l = j + 1;16             }17             for (int j = n - 1; j >= 0; j--) {18                 if (matrix[i][j] == '1') right[j] = min(right[j], r);19                 else right[j] = n, r = j;20             }21             for (int j = 0; j < n; j++)22                 area = max(area, (right[j] - left[j]) * height[j]);23         }24         return area;25     }26 };