Not very efficient algorithm since it does not use the provided index of a black pixel.

We know the m X n matrix is a valid rectangle that encloses all black pixels. We pin the upper left and lower right four indices and keep updating until a row or column does contain a '1'. This defines the smallest rectangle.

```
def minArea(self, image, x, y):
leftx = lefty = 0
rightx, righty = len(image) - 1, len(image[0]) - 1
for i in range(len(image)):
if '1' not in image[i]:
leftx += 1
else:
break
for j in range(len(image[0])):
if '1' not in [image[a][j] for a in range(len(image))]:
lefty += 1
else:
break
for m in range(len(image)-1, -1,-1):
if '1' not in image[m]:
rightx -= 1
else:
break
for n in range(len(image[0])-1,-1,-1):
if '1' not in [image[b][n] for b in range(len(image)-1,-1,-1)]:
righty -= 1
else:
break
return (rightx-leftx+1) * (righty-lefty+1)
```