Let's Code_: GeeksforGeeks Medium: Find the Number of Islands

Sunday, August 29, 2021

GeeksforGeeks Medium: Find the Number of Islands

Yesterday I solved a few hands-on coding problems using Java programming language.

I came across this easy problem(called Medium there) over GfG, to turn on the inertia:

Problem: Find the Number of Islands

Problem Description: https://practice.geeksforgeeks.org/problems/find-the-number-of-islands

Problem Approach: DFS

Time Complexity: O(V) where v = number of cells in the input grid

One Java Solution:

// { Driver Code Starts
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine().trim());
        while(T-->0)
        {
            String[] s = br.readLine().trim().split(" ");
            int n = Integer.parseInt(s[0]);
            int m = Integer.parseInt(s[1]);
            char[][] grid = new char[n][m];
            for(int i = 0; i < n; i++){
                String[] S = br.readLine().trim().split(" ");
                for(int j = 0; j < m; j++){
                    grid[i][j] = S[j].charAt(0);
                }
            }
            Solution obj = new Solution();
            int ans = obj.numIslands(grid);
            System.out.println(ans);
        }
    }
}// } Driver Code Ends



class Solution
{
    int r, c;
    byte[] xOffset = {1, 1, 1, 0, -1, -1, -1, 0};
    byte[] yOffset = {1, 0, -1, -1, -1, 0, 1, 1};
    
    //Function to find the number of islands.
    public int numIslands(char[][] grid)
    {
        // Code here
        r = grid.length;
        c = grid[0].length;
        
        boolean[][] visited = new boolean[r][c];
        int cnt = 0;
        for (int i=0; i<r; i++) {
            for(int j=0; j<c; j++) {
                if (grid[i][j]=='1' && !visited[i][j]) {
                    dfs(grid, visited, i, j);
                    cnt++;
                }
            }
        }
        
        return cnt;
    }
    
    private void dfs (char[][] grid, boolean[][] visited, int x, int y) {
        //input validation
        if (!valid (grid, x, y) || visited[x][y]==true) {
            return;
        }
        
        visited[x][y] = true;
        
        for (int i=0; i<8; i++) {
            int newX = x + xOffset[i];
            int newY = y + yOffset[i];
            
            if(valid(grid, newX, newY) && !visited[newX][newY]) {
                dfs(grid, visited, newX, newY);
            }
        }
    }
    
    boolean valid (char[][]grid, int x, int y) {
        if (x<0 || x>=r || y<0 || y>=c || grid[x][y] == '0'){
            return false;
        }
        return true;
    }
}