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Find all rectangles filled with 0

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:

# Python program to find all 
# rectangles filled with 0
  
def findend(i,j,a,output,index):
    x = len(a)
    y = len(a[0])
  
    # flag to check column edge case,
    # initializing with 0
    flagc = 0
  
    # flag to check row edge case,
    # initializing with 0
    flagr = 0
  
    for m in range(i,x): 
  
        # loop breaks where first 1 encounters
        if a[m][j] == 1
            flagr = 1 # set the flag
            break
  
        # pass because already processed
        if a[m][j] == 5
            pass
  
        for n in range(j, y): 
  
            # loop breaks where first 1 encounters
            if a[m][n] == 1:
                flagc = 1 # set the flag
                break
  
            # fill rectangle elements with any
            # number so that we can exclude
            # next time
            a[m][n] = 5
  
    if flagr == 1:
        output[index].append( m-1)
    else:
        # when end point touch the boundary
        output[index].append(m) 
  
    if flagc == 1:
        output[index].append(n-1)
    else:
        # when end point touch the boundary
        output[index].append(n) 
  
  
def get_rectangle_coordinates(a):
  
    # retrieving the column size of array
    size_of_array = len(a) 
  
    # output array where we are going
    # to store our output 
    output = [] 
  
    # It will be used for storing start
    # and end location in the same index
    index = -1
  
    for i in range(0,size_of_array):
        for j in range(0, len(a[0])):
            if a[i][j] == 0:
  
                # storing initial position 
                # of rectangle
                output.append([i, j]) 
  
                # will be used for the 
                # last position
                index = index + 1        
                findend(i, j, a, output, index) 
  
  
    print (output)
  
# driver code
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]]

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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This article is attributed to GeeksforGeeks.org

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