Python3 - Number of islands - LeetCode question

Question

I am trying to solve this leetcode question: https://leetcode.com/problems/number-of-islands/

    visited: List[List[int]] = None
    def numIslands(self, grid: List[List[str]]) -> int:
        self.visited = [[0] * len(grid[0])] * len(grid)
        m, n = 0, 0
        islands = 0
        while m < len(grid):
            while n < len(grid[0]):
                print(m,n)
                islands = self.recur(grid,m,n,islands)
                n = n + 1
            m = m +1
        return islands
        
    def recur(self, grid: List[List[str]],m: int, n: int, islands: int) -> int:
        print(m,n,grid[m][n])
        if grid[m][n] == "1" and self.visited[m][n] == 0:
            down, right = None, None
            if m < len(grid) - 1:
                down = self.recur(grid,m+1, n, islands)
            if n < len(grid[0]) - 1:
                right = self.recur(grid,m, n+1, islands)
            
            if right is not None and down is not None:
                islands = min(islands,down,right) + 1
            elif right is not None:
                islands = min(islands,right) + 1
            elif down is not None:
                islands = min(islands,down) + 1
            else:
                islands = islands + 1
        self.visited[m][n] = 1
        print(self.visited)
        return islands

My idea is to use recursion and if 1 is encountered keep going downwards and then towards right until a zero is encountered. I wanted to use a visited variable to keep track of all the already visited values in grid and skip recursion for those.

But I noticed that wrong cells are getting updated in visited :

for the first iteration, m =0 and n =0, here is the output:

0 0
0 0 1
1 0 1
2 0 1
3 0 0
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
2 1 1
3 1 0
[[1, 1, 0, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0]]
2 2 0
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
1 1 1
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
0 1 1
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]
[[1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0]]

All the 3 digits lines indicate the m , n and grid[m][n] values each time the recur method is called. I am printing the visited at the end of the method. I notice that the m , n values that were not visited were also getting updated to 1 in visited .

I am not able to figure out why this is happening and it's giving me a wrong result. Any help would be appreciated.

Answer 1

self.visited = [[0] * len(grid[0])] * len(grid) is going to give you issues. Initializing a 2D array like this causes references to the same mutable object (in this case [0] ) within your list items. In other words, modifying visited[0][0] changes all elements in visited[0] .

Try something like this:

visited = [['0'] * len(grid[0]) for _ in range(len(grid))]