711. Number of Distinct Islands II

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 they have the same shape, or have the same shape after rotation(90, 180, or 270 degrees only) or reflection(left/right direction or up/down direction).

Example 1:

11000
10000
00001
00011

Given the above grid map, return 1.

Notice that:

11
1

and

 1
11

are considered same island shapes. Because if we make a 180 degrees clockwise rotation on the first island, then two islands will have the same shapes.

Example 2:

11100
10001
01001
01110

Given the above grid map, return 2.

Here are the two distinct islands:

111
1

and

1
1

Notice that:

111
1

and

1
111

are considered the same island shapes. Because if we flip the first array in the up/down direction, then they have the same shapes.

Note:The length of each dimension in the givengriddoes not exceed 50.

Thoughts:

DFS + sorting: first use DFS to group all the islands, and then for "normalize" each group by listing all the possible

cases of index presentation of a island and pick the canonical one.

Code

class Solution {
    // typedef pair<int,int> point; old fashion :(
    using point = pair<int,int>; // new fashion :)
    using points = vector<point>;
    map<int, points> m;
public:
    int numDistinctIslands2(vector<vector<int>>& grid) {
        set <points> distinct; 
        if(grid.size()==0) return distinct.size();        
        // step1: find island
        int count = 1;
        for(int i = 0; i < grid.size(); i++)
            for(int j =0; j < grid[i].size(); j++)if(grid[i][j] == 1){
                    dfs(i,j,grid,++count);
                    distinct.insert(norm(m[count]));
                }
        return distinct.size();
    }

    void dfs(int r, int c, vector<vector<int>>&grid, int cnt){
        if (r < 0  || r >= grid.size() || c < 0 || c >= grid[0].size()) return;
        if(grid[r][c] != 1) return; // must be 1 since other number means "unpassable"
        grid[r][c] = cnt; // "infection" , "propogation"
        m[cnt].emplace_back(r,c);// record
        dfs(r + 1, c , grid, cnt); //down
        dfs(r - 1, c , grid, cnt); //up
        dfs(r , c + 1 , grid, cnt); //right
        dfs(r , c - 1, grid, cnt); //left
    }

    points norm(points v){
        // expand all the possibilities of a norm
        vector<points> spand(8);
        for(auto p: v){
            int x = p.first, y = p.second;
            // 8 cases
            spand[0].emplace_back(x , y);            
            spand[1].emplace_back(x , -y); // reflextion by vertical
            spand[2].emplace_back(-x , y); // reflextion by horizontal
            spand[3].emplace_back(-x , -y); // reflextion by (0,0)
            spand[4].emplace_back(y , -x); // rotation by 90 clockwise
            spand[5].emplace_back(-y , x); // rotation by 90 counter-clockwise
            spand[6].emplace_back(-y , -x); // rotation by 90 counter-clockwise + reflectin by vertical 
                                            // or rotation by 90 clockwise + reflection by horizontal
            spand[7].emplace_back(y , x);   // rotation by 90 clockwise + refletion by vertical 
                                            // rotation by 90 counter-clockwise + reflection by horizontal
        }

        for(auto &l: spand) sort(l.begin(), l.end());

        // normalization: fix the offset as (0,0)
        for (auto &l:spand) {
            for (int i = 1; i < v.size(); ++i) 
                l[i] = {l[i].first-l[0].first, l[i].second - l[0].second};
            l[0] = {0,0};
        }

        // finally sort the spand to get the canical representation
        sort(spand.begin(), spand.end());
        return spand[0];
    }

};