MaxAreaofIsland.cpp

/*
You are given an m x n binary matrix grid. An island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
Example 2:

Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0
 

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 50
grid[i][j] is either 0 or 1.

https://leetcode.com/explore/challenge/card/june-leetcoding-challenge-2021/603/week-1-june-1st-june-7th/3764/
*/

class Solution {
public:
    int sizeOfIsland(vector<vector<int>> &grid, vector<vector<int>> &visited, int i, int  j){
        if (i < 0 || (i > grid.size()-1) || j < 0 || (j > grid[0].size()-1) || grid[i][j] == 0 || visited[i][j] == 1){
            return 0;
        }
        visited[i][j] = 1;

        return (1 + sizeOfIsland(grid, visited, i+1, j) + sizeOfIsland(grid, visited, i-1, j) + sizeOfIsland(grid, visited, i, j+1) + sizeOfIsland(grid, visited, i, j-1));
        }
    
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int i, j, res=0;
        vector<vector<int>> visited(grid.size(), vector<int>(grid[0].size(), 0));
        for(i=0; i<grid.size(); i++){
            for(j=0; j<grid[0].size(); j++){
                if(visited[i][j]==0 || grid[i][j] == 1)
                    res = max(res, sizeOfIsland(grid, visited, i, j));
            }
        }
        return res;
    }
};