An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.
For example, given the following image:
[
“0010”,
“0110”,
“0100”
]
and x = 0, y = 2,
Return 6.
Hide Company Tags Google
Hide Tags Binary Search
Accept
Time complexity: O(mn)
class Solution(object):
def minArea(self, image, x, y):
""" :type image: List[List[str]] :type x: int :type y: int :rtype: int """
if len(image)==0 or len(image[0])==0 or x<0 or y<0:
return 0
res = [y, y, x, x]
dict_direct = [ [1,0], [-1,0], [0,1], [0,-1] ]
m,n = len(image), len(image[0])
visited = [[0]*n for dummy in range(m)]
self.helper( image, x, y, visited, res, dict_direct)
return (res[1]-res[0]+1)*(res[2]-res[3]+1)
def helper(self, image, cur_i, cur_j, visited, res, dict_direct):
visited[cur_i][cur_j] = 1
res[0], res[1], res[2], res[3] = \
min(res[0], cur_j), max(res[1], cur_j), \
max(res[2], cur_i), min(res[3], cur_i)
for ind in range(4):
next_i = cur_i + dict_direct[ind][0]
next_j = cur_j + dict_direct[ind][1]
if next_i<0 or next_i==len(image) \
or next_j<0 or next_j==len(image[0]) \
or image[next_i][next_j]=="0" \
or visited[next_i][next_j]==1:
continue
self.helper(image, next_i, next_j, visited, res, dict_direct)
return
Get idea from 1, 2, 3.