<p>Given an <code>m x n</code> 2D binary grid <code>grid</code> which represents a map of <code>&#39;1&#39;</code>s (land) and <code>&#39;0&#39;</code>s (water), return <em>the number of islands</em>.</p>

<p>An <strong>island</strong> 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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;]
]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;]
]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>grid[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

class Solution
{
public:
    void searchForConnectedLand(vector<vector<char>> &grid, int i, int j)
    {
        stack<pair<int, int>> landSquare;
        landSquare.push({i, j});

        // Search directions (up, down, left, right)
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        // Now we will check if the stack has any land('1')
        while (!landSquare.empty())
        {
            pair<int, int> currentSquare = landSquare.top();
            landSquare.pop();
            int x = currentSquare.first;
            int y = currentSquare.second;
            // We check first if we are out of bounds and if the coordinates are still an land.
            if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] != '1')
            {
                continue;
            }

            // Now we mark the square as visited(counted in the island)
            grid[x][y] = '0';

            // Now I add the squares around the current one to stack to be searched
            for (auto direction : directions)
            {
                int newSquareX = x + direction.first;
                int newSquareY = y + direction.second;

                landSquare.push({newSquareX, newSquareY});
            }
        }
    }

    int numIslands(vector<vector<char>> &grid)
    {
        if (grid.empty() || grid[0].empty())
        {
            return 0;
        }

        int numberOfIslands = 0;

        for (int i = 0; i < grid.size(); i++)
        {
            for (int j = 0; j < grid[0].size(); j++)
            {
                if (grid[i][j] == '1')
                {
                    numberOfIslands++;
                    searchForConnectedLand(grid, i, j);
                }
            }
        }

        return numberOfIslands;
    }
};
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