Python Program to Find all rectangles filled with 0
Last Updated :
28 Jul, 2022
We have one 2D array, filled with zeros and ones. We have to find the starting point and ending point of all rectangles filled with 0. It is given that rectangles are separated and do not touch each other however they can touch the boundary of the array.A rectangle might contain only one element.
Examples:
input = [
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1]
]
Output:
[
[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]
]
Explanation:
We have three rectangles here, starting from
(2, 3), (3, 1), (5, 3)
Input = [
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 0]
]
Output:
[
[0, 1, 0, 1], [1, 2, 1, 2], [2, 3, 3, 5],
[3, 1, 4, 1], [5, 3, 5, 6], [7, 2, 7, 2],
[7, 6, 7, 6]
]
Step 1: Look for the 0 row-wise and column-wise
Step 2: When you encounter any 0, save its position in output array and using loop change all related 0 with this position in any common number so that we can exclude it from processing next time.
Step 3: When you change all related 0 in Step 2, store last processed 0’s location in output array in the same index.
Step 4: Take Special care when you touch the edge, by not subtracting -1 because the loop has broken on the exact location.
Below is the implementation of above approach:
Python3
def findend(i,j,a,output,index):
x = len(a)
y = len(a[0])
flagc = 0
flagr = 0
for m in range(i,x):
if a[m][j] == 1:
flagr = 1
break
if a[m][j] == 5:
pass
for n in range(j, y):
if a[m][n] == 1:
flagc = 1
break
a[m][n] = 5
if flagr == 1:
output[index].append( m-1)
else:
output[index].append(m)
if flagc == 1:
output[index].append(n-1)
else:
output[index].append(n)
def get_rectangle_coordinates(a):
size_of_array = len(a)
output = []
index = -1
for i in range(0,size_of_array):
for j in range(0, len(a[0])):
if a[i][j] == 0:
output.append([i, j])
index = index + 1
findend(i, j, a, output, index)
print (output)
tests = [
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1]
]
get_rectangle_coordinates(tests)
|
Output:
[[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]]
Time Complexity: O(x*y).
Auxiliary Space: O(x*y).
Please refer complete article on Find all rectangles filled with 0 for more details!
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