694. Number of Distinct Islands (identical)

Given a non-empty 2D arraygridof 0's and 1's, an island is a group of1'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.

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 givengriddoes not exceed 50.

Thoughts:

1. Distinct islands: island 2d coordinates sets are distinct based off its offset

Code: Java

class Solution {
    private static final int d [] = {0,1,0,-1,0};

    public int numDistinctIslands(int[][] grid){
        int m = grid.length, n = grid[0].length;
        Set<List<List<Integer>>> distinctIslands = new HashSet<>();
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j++){
                List<List<Integer>> island = new ArrayList<>();
                if (dfs(i, j, i, j, grid, m, n, island)){
                    distinctIslands.add(island);
                }
            }
        }

        return distinctIslands.size();
    }

    private boolean dfs(int i, int j, int x, int y, int[][] grid, int m, int n, List<List<Integer>> island ){
        if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] <= 0) return false; // 1: island, -: visited path
        grid[x][y] = -1;
        island.add(Arrays.asList(x - i, y - j));
        for( int k = 0; k < 4; k ++){
            dfs(i, j, x + d[k], y + d[k + 1], grid, m, n, island);
        }

        return true;
    }
}

Code: C++

class Solution {
public:
    int numDistinctIslands(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        set<vector<vector<int>>> distinctIslands;
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j++){
                vector<vector<int>> island;
                if (dfs(i, j, i, j, grid, m, n, island)){
                    distinctIslands.insert(island);
                }
            }
        }
        return distinctIslands.size();
    }
private:
    int d [5]  = {0,1,0,-1,0};

    bool dfs(int i, int j, int x, int y, vector<vector<int>> & grid, int m, int n, vector<vector<int>> & island){
        if(x < 0 or x >= m or y < 0 or y >= n or grid[x][y] <= 0) return false;
        grid[x][y] *= -1;
        island.emplace_back(x-i, y-j);
        for(int k = 0; k < 4; k++){
            dfs(i, j, x + d[k], y + d[k + 1], grid, m, n, island);
        }

        return true;
    }

};