The DP solution proceeds row by row, starting from the first row. Let the maximal rectangle area at row i and column j be computed by [right(i,j) - left(i,j)]*height(i,j).

All the 3 variables left, right, and height can be determined by the information from previous row, and also information from the current row. So it can be regarded as a DP solution. The transition equations are:

left(i,j) = max(left(i-1,j), cur_left), cur_left can be determined from the current row

right(i,j) = min(right(i-1,j), cur_right), cur_right can be determined from the current row

height(i,j) = height(i-1,j) + 1, if matrix[i][j]=='1';

height(i,j) = 0, if matrix[i][j]=='0'

The code is as below. The loops can be combined for speed but I separate them for more clarity of the algorithm.

```
class Solution {public:
int maximalRectangle(vector<vector<char> > &matrix) {
if(matrix.empty()) return 0;
const int m = matrix.size();
const int n = matrix[0].size();
int left[n], right[n], height[n];
fill_n(left,n,0); fill_n(right,n,n); fill_n(height,n,0);
int maxA = 0;
for(int i=0; i<m; i++) {
int cur_left=0, cur_right=n;
for(int j=0; j<n; j++) { // compute height (can do this from either side)
if(matrix[i][j]=='1') height[j]++;
else height[j]=0;
}
for(int j=0; j<n; j++) { // compute left (from left to right)
if(matrix[i][j]=='1') left[j]=max(left[j],cur_left);
else {left[j]=0; cur_left=j+1;}
}
// compute right (from right to left)
for(int j=n-1; j>=0; j--) {
if(matrix[i][j]=='1') right[j]=min(right[j],cur_right);
else {right[j]=n; cur_right=j;}
}
// compute the area of rectangle (can do this from either side)
for(int j=0; j<n; j++)
maxA = max(maxA,(right[j]-left[j])*height[j]);
}
return maxA;
}
```

};

If you think this algorithm is not easy to understand, you can try this example:

```
0 0 0 1 0 0 0
0 0 1 1 1 0 0
0 1 1 1 1 1 0
```

The vector "left" and "right" from row 0 to row 2 are as follows

row 0:

```
l: 0 0 0 3 0 0 0
r: 7 7 7 4 7 7 7
```

row 1:

```
l: 0 0 2 3 2 0 0
r: 7 7 5 4 5 7 7
```

row 2:

```
l: 0 1 2 3 2 1 0
r: 7 6 5 4 5 6 7
```

The vector "left" is computing the left boundary. Take (i,j)=(1,3) for example. On current row 1, the left boundary is at j=2. However, because matrix[1][3] is 1, you need to consider the left boundary on previous row as well, which is 3. So the real left boundary at (1,3) is 3.

I hope this additional explanation makes things clearer.