DFS: Connected Cell in a Grid

Sort by

recency

|

232 Discussions

|

  • + 0 comments

    C# DFS

        public static int maxRegion(List<List<int>> grid)
        {
            int res = 0;
            int n = grid.Count();
            int m = grid[0].Count();
            if(n == 0 || m == 0) return res;
            int [,] visited = new int[n,m];
            List<List<int>> directions = new List<List<int>>{
                new List<int>{0, 1},
                new List<int>{0, -1},
                new List<int>{-1, 0},
                new List<int>{1, 0},
                new List<int>{1, 1},
                new List<int>{-1, -1},
                new List<int>{-1, 1},
                new List<int>{1, -1},
            };
            int area = 0;
            void dfs(int row, int col){
                if(row < 0 || row >= n || col < 0 || col >= m) return;
                if(grid[row][col] == 0 || visited[row, col] == 1) return;
                area++;
                visited[row, col] = 1;
                foreach(var dir in directions)
                {
                    dfs(row+dir[0], col+dir[1]);
                }
            }
            for(int i = 0; i < n; i++)
            {
                for(int j = 0; j < m; j++)
                {
                    if(visited[i,j] == 0)
                    {
                        area = 0;
                        dfs(i, j);
                        res = Math.Max(res, area);
                    }
                }
            }
            return res;
        }
    
  • + 0 comments

    why do they want DFS? component size problems are more intuitively done with BFS.

    DFS recursive, not iterative:

    vector<int> S = {-1, 0, 1};
    
    void component(int i, int j, int n, int m, const vector<vector<int>>& grid, vector<vector<bool>>& visited, int& size) {
        visited[i][j] = true;
        ++size;
        for (int I : S) {
            for (int J : S) {
                if (i+I < 0 or i+I >= n or j+J < 0 or j+J >= m or visited[i+I][j+J] or grid[i+I][j+J] == 0) continue;
                component(i+I, j+J, n, m, grid, visited, size);
            }
        }
    }
    
    int maxRegion(int n, int m, const vector<vector<int>>& grid) {
        vector<int> components;
        vector<vector<bool>> visited(n, vector<bool>(m));
        int size = 0;
        for (int i=0; i < n; ++i) {
            for (int j=0; j < m; ++j) {
                if (visited[i][j] or grid[i][j] == 0) continue;
                size = 0;
                component(i, j, n, m, grid, visited, size);
                components.push_back(size);
            }
        }
        return *max_element(components.begin(), components.end());
    }
    
    int main()
    {
        int n, m, k;
        cin >> n >> m;
        vector<vector<int>> grid(n, vector<int>(m));
        for (int i=0; i < n; ++i) {
            for (int j=0; j < m; ++j) {
                cin >> k;
                grid[i][j] = k;
            }
        }
        cout << maxRegion(n, m, grid);
    }
    
  • + 1 comment

    This was my timed solution. I am not very good a t matrix problems so if anyone could view my code and tell me why it may be abd or ugly or any improvements I'd greatly appreciate it!

    int maxRegion(vector> grid) { int max_region_amt = 0; int n = grid.size(); int m = grid[0].size(); stack q; vector visited(n*m); int i = 0; while(i < n * m){

        if(grid[i / m][i % m]){
            break;
        }
         visited[i] = true;
        ++i;
    }
    
    while (i < n*m && !visited[i]) {
        visited[i] = true;
        int amt_region = 1;
        q.push(i);
        cout << "New iteration: " << i << "\n";
        while(q.size()){
            int curr_value = q.top();
            q.pop();
            if(curr_value / m != 0){
                int prev_row =(curr_value - m)/m;
                if(!visited[curr_value - m]){
                    visited[curr_value - m] = true;
                    if(grid[prev_row][curr_value % m]){
                            q.push(curr_value - m);
                            amt_region++;
                        }
                }
                if(curr_value % m && !visited[curr_value - m - 1]){
                    visited[curr_value - m - 1] = true;;
                    if(grid[prev_row][curr_value % m -1]){
                        q.push(curr_value - m - 1);
                        amt_region++;
                    }
                }
                if((curr_value + 1) % m && !visited[curr_value - m + 1]){
                    visited[curr_value - m + 1] = true;;
                    if(grid[prev_row][curr_value % m +1]){
                        q.push(curr_value - m + 1);
                        amt_region++;
                    }
                }
            }
            if(curr_value / m + 1 != n){
                int next_row =(curr_value + m)/m;
                if(!visited[curr_value + m]){
                    visited[curr_value + m] = true;
                    if(grid[next_row][curr_value % m]){
                            q.push(curr_value + m);
                            amt_region++;
                        }
                }
                if(curr_value % m && !visited[curr_value + m - 1]){
                    visited[curr_value + m - 1] = true;;
                    if(grid[next_row][curr_value % m - 1]){
                        q.push(curr_value + m - 1);
                        amt_region++;
                    }
                }
                if((curr_value + 1) % m && !visited[curr_value + m + 1]){
                    visited[curr_value + m + 1] = true;;
                    if(grid[next_row][curr_value % m +1]){
                        q.push(curr_value + m + 1);
                        amt_region++;
                    }
                }
            }
            if((curr_value + 1) % m && !visited[curr_value + 1]){
                visited[curr_value + 1] = true;;
                if(grid[curr_value / m][curr_value % m +1]){
                    q.push(curr_value + 1);
                    amt_region++;
                }
            }
            if((curr_value) % m && !visited[curr_value - 1]){
                visited[curr_value - 1] = true;
                if(grid[curr_value / m][curr_value % m - 1]){
                    q.push(curr_value +-1);
                    amt_region++;
                }
            }
    
            while(i < n * m && (visited[i] || !grid[i / m][i % m])){
                visited[i] = true;
                ++i;
            }
        }
        max_region_amt = max(max_region_amt, amt_region);
    }
    
    
    
    
    cout << i << "\n";
    return max_region_amt;
    

    }

  • + 0 comments

    All the graph methods are reasonable and make sense given the category

    However, is it possible to solve this with a couple convolutional passes?

  • + 0 comments

    Python

    def neighborhood(i, j):
        return [(i+1, j), (i-1, j), (i, j+1), (i, j-1), (i+1, j+1), (i-1, j-1), (i-1, j+1), (i+1, j-1)]
    
    def maxRegion(grid):
        largest_region_size = 0
        visited = set()
        for i in range(len(grid)-1):
            for j in range(len(grid[0])):
                if (i, j) in visited or grid[i][j] == 0:
                    continue
                region_size = 1
                visited.add((i, j))
                stack = neighborhood(i, j)
                while stack:
                    curr = stack.pop()
                    if curr in visited or curr[0] < 0 or curr[1] < 0:
                        continue
                    try:
                        if grid[curr[0]][curr[1]] == 0:
                            continue
                    except:
                        continue
                    visited.add((curr[0], curr[1]))
                    region_size += 1
                    stack.extend(neighborhood(curr[0], curr[1]))
                largest_region_size = max(largest_region_size, region_size)
                    
                
        return largest_region_size