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[Matrix] Max area of island problem solved DFS algorithm (TheAlgorith…
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* Maximum area of island program added

* Update matrix/max_area_of_island.py

Co-authored-by: Caeden <caedenperelliharris@gmail.com>

* Update matrix/max_area_of_island.py

Co-authored-by: Caeden <caedenperelliharris@gmail.com>

* Update matrix/max_area_of_island.py

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* max area of island

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* Update matrix/max_area_of_island.py

Co-authored-by: Caeden Perelli-Harris <caedenperelliharris@gmail.com>

* Update matrix/max_area_of_island.py

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* Update matrix/max_area_of_island.py

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"""
Given an two dimensional binary matrix grid. An island is a group of 1's (representing
land) connected 4-directionally (horizontal or vertical.) You may assume all four edges
of the grid are surrounded by water. The area of an island is the number of cells with
a value 1 in the island. Return the maximum area of an island in a grid. If there is no
island, return 0.
"""

matrix = [
[0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
[0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0],
[0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
]


def is_safe(row: int, col: int, rows: int, cols: int) -> bool:
"""
Checking whether coordinate (row, col) is valid or not.
>>> is_safe(0, 0, 5, 5)
True
>>> is_safe(-1,-1, 5, 5)
False
"""
return 0 <= row < rows and 0 <= col < cols


def depth_first_search(row: int, col: int, seen: set, mat: list[list[int]]) -> int:
"""
Returns the current area of the island
>>> depth_first_search(0, 0, set(), matrix)
0
"""
rows = len(mat)
cols = len(mat[0])
if is_safe(row, col, rows, cols) and (row, col) not in seen and mat[row][col] == 1:
seen.add((row, col))
return (
1
+ depth_first_search(row + 1, col, seen, mat)
+ depth_first_search(row - 1, col, seen, mat)
+ depth_first_search(row, col + 1, seen, mat)
+ depth_first_search(row, col - 1, seen, mat)
)
else:
return 0


def find_max_area(mat: list[list[int]]) -> int:
"""
Finds the area of all islands and returns the maximum area.
>>> find_max_area(matrix)
6
"""
seen: set = set()

max_area = 0
for row, line in enumerate(mat):
for col, item in enumerate(line):
if item == 1 and (row, col) not in seen:
# Maximizing the area
max_area = max(max_area, depth_first_search(row, col, seen, mat))
return max_area


if __name__ == "__main__":
import doctest

print(find_max_area(matrix)) # Output -> 6

"""
Explanation:
We are allowed to move in four directions (horizontal or vertical) so the possible
in a matrix if we are at x and y position the possible moving are
Directions are [(x, y+1), (x, y-1), (x+1, y), (x-1, y)] but we need to take care of
boundary cases as well which are x and y can not be smaller than 0 and greater than
the number of rows and columns respectively.
Visualization
mat = [
[0,0,A,0,0,0,0,B,0,0,0,0,0],
[0,0,0,0,0,0,0,B,B,B,0,0,0],
[0,C,C,0,D,0,0,0,0,0,0,0,0],
[0,C,0,0,D,D,0,0,E,0,E,0,0],
[0,C,0,0,D,D,0,0,E,E,E,0,0],
[0,0,0,0,0,0,0,0,0,0,E,0,0],
[0,0,0,0,0,0,0,F,F,F,0,0,0],
[0,0,0,0,0,0,0,F,F,0,0,0,0]
]
For visualization, I have defined the connected island with letters
by observation, we can see that
A island is of area 1
B island is of area 4
C island is of area 4
D island is of area 5
E island is of area 6 and
F island is of area 5
it has 6 unique islands of mentioned areas
and the maximum of all of them is 6 so we return 6.
"""

doctest.testmod()

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