<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>

<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,0]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1],[1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class DisJointSet:
    def __init__(self, n):
        self.rank = [0 for i in range(n+1)]
        self.size = [1 for i in range(n+1)]
        self.parent = [i for i in range(n+1)]
        
    def findUPar(self,u):
        if u == self.parent[u]:
            return u 
        self.parent[u] = self.findUPar(self.parent[u])
        return self.parent[u]
        
    def unionByRank(self,u,v):
        parentA = self.findUPar(u)
        parentB = self.findUPar(v)
        
        if parentA == parentB:
            return
        
        if self.rank[parentA] < self.rank[parentB]:
            self.parent[parentA] = parentB
        elif self.rank[parentB] < self.rank[parentA]:
            self.parent[parentB] = parentA
        else:
            self.parent[parentB] = parentA
            self.rank[parentA] += 1

    def unionBySize(self,u,v):
        parentA = self.findUPar(u)
        parentB = self.findUPar(v)

        if parentA == parentB:
            return
        
        if self.size[parentA] < self.size[parentB]:
            self.parent[parentA] = parentB
            self.size[parentB] += self.size[parentA]
        else:
            self.parent[parentB] = parentA
            self.size[parentA] += self.size[parentB]
       
class Solution:
    def isValid(self, ROWS, COLS, x, y):
        return x >= 0 and x < ROWS and y >= 0 and y < COLS

    def largestIsland(self, grid: List[List[int]]) -> int:
        n,m = len(grid), len(grid[0])
        # visited = [[0 for i in range(m)] for r in range(n)]
        
        ds = DisJointSet(n*m)
        
        directions = [-1,0,1,0,-1]

        countZeros = 0
        for r in range(n):
            for c in range(m):
                if grid[r][c] == 0:
                    countZeros += 1
                    continue
                for d in range(4):
                    newR = r + directions[d]
                    newC = c + directions[d+1]

                    if self.isValid(n,m,newR,newC) and grid[newR][newC] == 1:
                        currNode = n*r + c
                        adjNode = n*newR + newC
                        if ds.findUPar(currNode) != ds.findUPar(adjNode):
                            ds.unionBySize( currNode, adjNode)

        if countZeros == 0:
            return n*m
            
        ans = 0
        for r in range(n):
            for c in range(m):
                if grid[r][c] == 1:
                    continue
                
                curr = 1
                visitedDict = dict()

                for d in range(4):
                    newR = r + directions[d]
                    newC = c + directions[d+1]
                    if self.isValid(n,m,newR,newC) and grid[newR][newC] == 1:
                        adjNode = n*newR + newC
                        par = ds.findUPar(adjNode)
                        if visitedDict.get(par,0) == 0:
                            size = ds.size[par]
                            curr += size
                            visitedDict[par] = 1
                
                ans = max(ans, curr)
        
        return ans
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