NumberOfIslands.h

// Leetcode 200, number of islands.
/*
Given an m x n 2D binary grid grid which represents a map of '1' s(land) and '0' s(water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally
or vertically.You may assume all four edges of the grid are all surrounded by water.
Example 1 :

    Input : grid = [
    [ "1", "1", "1", "1", "0" ],
    [ "1", "1", "0", "1", "0" ],
    [ "1", "1", "0", "0", "0" ],
    [ "0", "0", "0", "0", "0" ]
] Output : 1 Example 2 :

    Input : grid = [
        [ "1", "1", "0", "0", "0" ],
        [ "1", "1", "0", "0", "0" ],
        [ "0", "0", "1", "0", "0" ],
        [ "0", "0", "0", "1", "1" ]
    ] Output : 3

                   Constraints :

    m
    == grid.length n
    == grid[i].length 1 <= m,
            n <= 300 grid[i][j] is '0' or '1'.
*/

#include <vector>
#include "headers.h"

class Solution
{
public:

    void mark(int i, int j, vector<vector<char>> &grid)
    {
        if(i<0 || i>= grid.size() || j < 0 || j >= grid[0].size())
        {
            return;
        }

        if(grid[i][j] ==0)
        {
            return;
        }

        grid[i][j] = 0;
        mark(i - 1, j, grid);
        mark(i + 1, j, grid);
        mark(i , j - 1, grid);
        mark(i , j+1, grid);
    }

    int numIslands(vector<vector<char>> &grid)
    {
        int row = grid.size();
        if(row == 0)
        {
            return 0;
        }

        int col = grid[0].size();
        if(col == 0)
        {
            return 0;
        }

        int nIslands = 0;
        for (int i = 0; i < row;i++)
        {
            for (int j = 0; j < col; j++)
            {
                if(grid[i][j] !=0 )
                {
                    nIslands++;
                    mark(i, j, grid);
                }
            }
        }
    }
};