<p>You are given an <code>m x n</code> binary matrix <code>grid</code>. An island is a group of <code>1</code>&#39;s (representing land) connected <strong>4-directionally</strong> (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.</p>

<p>The <strong>area</strong> of an island is the number of cells with a value <code>1</code> in the island.</p>

<p>Return <em>the maximum <strong>area</strong> of an island in </em><code>grid</code>. If there is no island, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg" style="width: 500px; height: 310px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The answer is not 11, because the island must be connected 4-directionally.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[0,0,0,0,0,0,0,0]]
<strong>Output:</strong> 0
</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;= 50</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class DisjointSet {
    public:
        vector<int> parent;
        vector<int> size;

        DisjointSet(int n) {
            parent.resize(n);
            size.resize(n, 1);
            for (int i = 0; i < n; i++) parent[i] = i;
        }

        int findUPar(int node) {
            if (node == parent[node]) return node;
            return parent[node] = findUPar(parent[node]);
        }

        void unionBySize(int n1, int n2) {
            int p1 = findUPar(n1);
            int p2 = findUPar(n2);

            if (p1 == p2) return;

            if (size[p1] > size[p2]) {
                size[p1] += size[p2];
                parent[p2] = p1;
            }
            else {
                size[p2] += size[p1];
                parent[p1] = p2;
            }
            return;
        }
};

class Solution {
public:
    void traverse(vector<vector<int>>& grid, vector<vector<bool>>& visited, DisjointSet& ds, int row, int col) {
        queue<pair<int,int>> q;
        q.push({row, col});
        int rows = grid.size();
        int cols = grid[0].size();
        while (!q.empty()) {
            int ci = q.front().first;
            int cj = q.front().second;
            q.pop();
            for (int di = -1; di <= 1; di++) {
                for (int dj = -1; dj <= 1; dj++) {
                    if (di != 0 && dj != 0) continue;

                    int ni = ci + di;
                    int nj = cj + dj;

                    if (ni < 0 || nj < 0 || ni == rows || nj == cols) continue;

                    if (!visited[ni][nj] && grid[ni][nj] == 1) {
                        ds.unionBySize(cols * ni + nj, cols * ci + cj);
                        visited[ni][nj] = true;
                        q.push({ni, nj});
                    }
                }
            }
        }
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {

        int rows = grid.size();
        int cols = grid[0].size();
        DisjointSet ds(rows * cols);
        vector<vector<bool>> visited (rows, vector<bool> (cols));
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (!visited[i][j] && grid[i][j] == 1) traverse(grid, visited, ds, i, j);
            }
        }
        int ans = 0;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1 && ds.findUPar(cols * i + j) == cols * i + j) ans = max(ans, ds.size[cols * i + j]);
            }
        }
        return ans;
    }
};
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