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Program to find number of distinct island shapes from a given matrix in Python
Suppose we have a 2d binary matrix, we have to find the number of distinct islands in the given matrix. Here 1 represents land and 0 represents water, so an island is a set of 1s that are close and whose perimeter is surrounded by water. Here two islands are unique if their shapes are different.
So, if the input is like
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 |
then the output will be 4 (distinct islands are in different color).
To solve this, we will follow these steps −
Define a function dfs() . This will take i, j, k, l
mat[i, j] := 0
insert pair (i − k, j − l) at the end of shape
if i + 1 < row count of mat and mat[i + 1, j] is 1, then
dfs(i + 1, j, k, l)
if j + 1 < column count of mat and mat[i, j + 1] is 1, then
dfs(i, j + 1, k, l)
if i − 1 >= 0 and mat[i − 1, j] is 1, then
dfs(i − 1, j, k, l)
if j − 1 >= 0 and mat[i, j − 1] is 1, then
dfs(i, j − 1, k, l)
From the main method do the following −
cnt := 0
shapes := a new set
for i in range 0 to row count of mat, do
for j in range 0 to column count of mat, do
if mat[i, j] is 1, then
shape := a new list
dfs(i, j, i, j)
if shape is not in shapes, then
cnt := cnt + 1
insert shape into shapes
return cnt
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, mat): def dfs(i, j, k, l): mat[i][j] = 0 shape.append((i − k, j − l)) if i + 1 < len(mat) and mat[i + 1][j]: dfs(i + 1, j, k, l) if j + 1 < len(mat[0]) and mat[i][j + 1]: dfs(i, j + 1, k, l) if i − 1 >= 0 and mat[i − 1][j]: dfs(i − 1, j, k, l) if j − 1 >= 0 and mat[i][j − 1]: dfs(i, j − 1, k, l) cnt = 0 shapes = set() for i in range(len(mat)): for j in range(len(mat[0])): if mat[i][j]: shape = [] dfs(i, j, i, j) shape = tuple(shape) if shape not in shapes: cnt += 1 shapes.add(shape) return cnt ob = Solution() matrix = [ [1, 0, 0, 0, 0], [1, 0, 1, 0, 1], [0, 1, 1, 0, 1], [0, 0, 1, 0, 0], [1, 0, 0, 0, 0], [1, 1, 0, 1, 1] ] print(ob.solve(matrix))
Input
[ [1, 0, 0, 0, 0], [1, 0, 1, 0, 1], [0, 1, 1, 0, 1], [0, 0, 1, 0, 0], [1, 0, 0, 0, 0], [1, 1, 0, 1, 1] ]
Output
4
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