<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>

struct Index
{
    int x = -1, y = -1;
};

class Solution
{
    void push_to_stack(
        const vector<vector<char>> &grid,
        vector<vector<int>> &nodeStates,
        std::vector<Index> &dfsStack,
        int X,
        int Y)
    {
        if (grid[X][Y] == '1' && nodeStates[X][Y] == 0)
        {
            nodeStates[X][Y] = 1;
            dfsStack.push_back({X, Y});
        }
    }
public:
    int numIslands(vector<vector<char>> &grid)
    {
        int islands = 0;
        // To keep track of the node traversal state
        // 0: unexplored
        // 1: visited
        // 2: completed
        vector<vector<int>> nodeStates;
        for (auto i = 0ul; i < grid.size(); ++i)
        {
            nodeStates.push_back(vector<int>());
            for (auto j = 0ul; j < grid[i].size(); ++j)
            {
                nodeStates[i].push_back(0);
            }
        }
        // DFS traversal stack stores node index and next neighbor index
        // to be explored of the node
        std::vector<Index> dfsStack;
        dfsStack.reserve(10000);

        for (int i = 0; i < grid.size(); ++i)
        {
            for (int j = 0; j < grid[i].size(); ++j)
            {
                if (grid[i][j] == '1' && nodeStates[i][j] == 0)
                {
                    islands += 1;

                    // Start a DFS search
                    nodeStates[i][j] = 1;
                    dfsStack.push_back({i, j});

                    while (!dfsStack.empty())
                    {
                        auto tos = dfsStack.back();
                        dfsStack.pop_back();
                        auto X = tos.x;
                        auto Y = tos.y;
                        if(X - 1 >= 0) 
                            push_to_stack(grid, nodeStates, dfsStack, X-1, Y);
                        if(X + 1 < grid.size()) 
                            push_to_stack(grid, nodeStates, dfsStack, X+1, Y);
                        if(Y - 1 >= 0) 
                            push_to_stack(grid, nodeStates, dfsStack, X, Y-1);
                        if(Y + 1 < grid[0].size()) 
                            push_to_stack(grid, nodeStates, dfsStack, X, Y+1);
                        nodeStates[X][Y] = 2;
                        
                    }
                }
            }
        }

        return islands;
    }
};
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