<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 DS{
public:
    vector<int> parent, size;
    DS(int n){
        parent.resize(n+1);
        iota(parent.begin(), parent.end(), 0);
        size.resize(n+1, 1);
    }

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

    void merge(int u, int v){
        int upar = findPar(u);
        int vpar = findPar(v);

        if(upar == vpar) return;

        if(size[upar] < size[vpar]){
            parent[upar] = vpar;
            size[vpar] += size[upar];
        }
        else{
            parent[vpar] = upar;
            size[upar] += size[vpar];
        }
    }
};

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        DS ds(m*n);

        int dx4[4] = {0, 0, -1, 1};
        int dy4[4] = {-1, 1, 0, 0};

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    int nodeV = (i*n) + j;

                    for(int k = 0; k < 4; k++){
                        int ni = i + dx4[k];
                        int nj = j + dy4[k];
                        if(ni >= 0 && nj >= 0 && ni < m && nj < n && grid[ni][nj]){
                            int adjN = (ni*n) + nj;
                            ds.merge(nodeV, adjN);
                        }
                    }
                }
            }
        }

        int maxSz = 0;
        for(int i = 0; i < m*n; i++){
            int c = i % n;
            int r = i / n;
            if(ds.parent[i] == i && grid[r][c] == 1){
                maxSz = max(ds.size[i], maxSz);
            }
        }

        return maxSz;
    }
};
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