<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 Solution:

    def markAllIslandsGetWaters(self,grid,directions):
        # dfs , mark from 2 , save size 
        m , n = len(grid),len(grid[0])
        waters = []
        island_diction = {}

        def dfs(i,j,island_index):
            grid[i][j] = island_index  # mark cell with island index
            count = 1  # mark 1 for each cell 

            for d in directions: 
                next_i, next_j = d[0] +i , d[1]+j
                if 0<=next_i<m and 0<=next_j<n and  grid[next_i][next_j]==1: 
                    count += dfs(next_i, next_j, island_index) # add comming back 
        
            return count 
     

        island_index = 2 # island index start from 2 

        for i in range(m): 
            for j in range(n): 
                if grid[i][j] ==1: 
                    island_size = dfs(i,j,island_index)
                    #print('count: ',island_size)
                    island_diction[island_index] = island_size # track island size
                    island_index +=1 

                elif grid[i][j] == 0: 
                    waters.append((i,j))

        return waters ,island_diction
    

    def findLargestIsland(self,waters,island_diction, grid,directions): 
        

        m ,n = len(grid), len(grid[0])
        largest = 0
     
        for i, j in waters: 
            current_largest = 0 # set current largest / reset 
            seen_islands = set()
            
            for d in directions: 
                next_i,next_j = d[0]+i,d[1]+j

                if 0<=next_i<m and 0<=next_j<n and grid[next_i][next_j]!=0:  # not water
                    curr_island_index =grid[next_i][next_j]

                    if curr_island_index not in seen_islands: 
                        seen_islands.add(curr_island_index) # keep visited islands
                        current_largest+=island_diction[curr_island_index] # add island size 
                
            largest = max(current_largest,largest)

        return largest +1

    def largestIsland(self, grid: List[List[int]]) -> int:
        directions = [(-1,0),(1,0),(0,-1),(0,1)]
        waters, islands_diction = self.markAllIslandsGetWaters(grid, directions) # list[tuple] (i,j)
        # print(islands_diction)
        # print('waters: ',waters)
        if not waters: 
          for k in islands_diction: 
            return islands_diction[k]

        return self.findLargestIsland(waters,islands_diction,grid,directions)





        
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