<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:
    void bfs(int row, int col, vector<vector<bool>> &visited, vector<vector<char>>& grid) {
        // mark the node as the visited
        visited[row][col] = true;

        // m rows, n columns
        int m = grid.size(), n = grid[0].size();

        // queue for BFS traversal
        queue<pair<int, int>> q;
        q.push({row, col});

        while(!q.empty()) {
            int curr_row = q.front().first;
            int curr_col = q.front().second;
            q.pop();

            // directions in which row & col neighbours to find
            vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

            for(auto it: directions) {
                // neighbour row & col indices
                int new_row = curr_row + it.first;
                int new_col = curr_col + it.second;
                
                if (new_row >=0 && new_row < m          // if row index is valid
                    && new_col >=0 && new_col < n       // if col index is valid
                    && grid[new_row][new_col] == '1'    // if node is island
                    && !visited[new_row][new_col]) {    // if node is not visited

                    // mark the node as visited & add to queue
                    visited[new_row][new_col] = true;
                    q.push({new_row, new_col});
                }
            }
        }
    }

    int numIslands(vector<vector<char>>& grid) {
        // m rows, n columns
        int m = grid.size(), n = grid[0].size();
        
        // visited matrix for each node in grid
        vector<vector<bool>> visited(m, vector<bool> (n, false));

        // number of starting nodes for traversal
        int cnt = 0;

        // for each node
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(!visited[i][j] && grid[i][j] == '1') {
                    cnt++;
                    bfs(i, j, visited, grid);
                }
            }
        }

        return cnt;
    }
};
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