Leetcode 694. Number of Distinct Islands (Python)

Related Topic

Depth-First-Search.

Description

Given a non-empty 2D array grid of 0’s and 1’s, 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.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Sample I/O

Example 1

11000
11000
00011
00011

Given the above grid map, return 1.

Example 2

11011
10000
00001
11011

Given the above grid map, return 3.

Notice That:

11
1

and

 1
11

are considered different island shapes, because we do not consider reflection / rotation.

Note:

The length of each dimension in the given grid does not exceed 50.

Methodology

This question can be solved by Depth First Search and It is similar with question 200. Number of Islands.The difference if to find the number of distinct islands

To find the number of distince islands, we need to record the shape of each island. If two islands with same shape then they are not distinct and count as 1 island. To find the island, we will use dfs and it will be same as question 200 to find the number of islands.

Code

def numDistinctIslands(self, grid: List[List[int]]) -> int:
        row = len(grid)
        col = len(grid[0])
        shapes = set()
        directions = [(-1,0),(0,1),(1,0),(0,-1)]
        visited=set()
        def dfs(x, y, pos, island_direction):
            for dx, dy in directions:
                nx,ny=x+dx,y+dy
                if 0<=nx <row and 0<= ny <col and grid[nx][ny] and (nx,ny) not in visited:
                    temp_direction = (pos[0]+dx, pos[1]+dy)
                    visited.add((nx,ny))
                    island_direction.append(temp_direction)
                    dfs(nx, ny, temp_direction,island_direction)
            return tuple(island_direction)
        
        for x in range(row):
            for y in range(col):
                if grid[x][y] and (x,y) not in visited:
                    visited.add((x,y))
                    shapes.add(dfs(x,y,(0,0),[(0,0)]))
        return len(shapes)

BigO

We iterate each cell will cost O(m*n) where m*n is the size of the 2D list. Each cell will also iterate four directions, so in total is O(m*n*4)