<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>

class Solution {
public:
// Converts a 2D grid of chars to an adjacency list of node indices
vector<vector<int>> convertGridTOAdjList(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> adjList(m * n); // adjacency list

        auto toIndex = [&](int r, int c) {
            return r * n + c;
        };

        // Directions for the 4 possible movements: up, down, left, right
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') { // Only consider land cells
                    int currentIndex = toIndex(i, j);
                    for (auto& dir : directions) {
                        int ni = i + dir.first;
                        int nj = j + dir.second;
                        if (ni >= 0 && ni < m && nj >= 0 && nj < n && grid[ni][nj] == '1') {
                            int neighborIndex = toIndex(ni, nj);
                            adjList[currentIndex].push_back(neighborIndex);
                        }
                    }
                }
            }
        }

        return adjList;
    }

    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<bool> visited(m * n, false);
        vector<vector<int>> adj = convertGridTOAdjList(grid);
        int count = 0;

        for (int i = 0; i < m * n; ++i) {
            if (!visited[i] && grid[i / n][i % n] == '1') {
                count++;
                dfs(i, adj, visited);
            }
        }
        
        return count;
    }

    void dfs(int start, const vector<vector<int>>& adj, vector<bool>& visited) {
        stack<int> st;
        st.push(start);
        visited[start] = true;

        while (!st.empty()) {
            int curr = st.top();
            st.pop();

            for (int i : adj[curr]) {
                if (!visited[i]) {
                    visited[i] = true;
                    st.push(i);
                }
            }
        }
    }
};

33264