<p>There is a <strong>1-based</strong> binary matrix where <code>0</code> represents land and <code>1</code> represents water. You are given integers <code>row</code> and <code>col</code> representing the number of rows and columns in the matrix, respectively.</p>

<p>Initially on day <code>0</code>, the <strong>entire</strong> matrix is <strong>land</strong>. However, each day a new cell becomes flooded with <strong>water</strong>. You are given a <strong>1-based</strong> 2D array <code>cells</code>, where <code>cells[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> represents that on the <code>i<sup>th</sup></code> day, the cell on the <code>r<sub>i</sub><sup>th</sup></code> row and <code>c<sub>i</sub><sup>th</sup></code> column (<strong>1-based</strong> coordinates) will be covered with <strong>water</strong> (i.e., changed to <code>1</code>).</p>

<p>You want to find the <strong>last</strong> day that it is possible to walk from the <strong>top</strong> to the <strong>bottom</strong> by only walking on land cells. You can start from <strong>any</strong> cell in the top row and end at <strong>any</strong> cell in the bottom row. You can only travel in the<strong> four</strong> cardinal directions (left, right, up, and down).</p>

<p>Return <em>the <strong>last</strong> day where it is possible to walk from the <strong>top</strong> to the <strong>bottom</strong> by only walking on land cells</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/1.png" style="width: 624px; height: 162px;" />
<pre>
<strong>Input:</strong> row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/2.png" style="width: 504px; height: 178px;" />
<pre>
<strong>Input:</strong> row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 1.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/3.png" style="width: 666px; height: 167px;" />
<pre>
<strong>Input:</strong> row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 3.
</pre>

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

<ul>
	<li><code>2 &lt;= row, col &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>4 &lt;= row * col &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>cells.length == row * col</code></li>
	<li><code>1 &lt;= r<sub>i</sub> &lt;= row</code></li>
	<li><code>1 &lt;= c<sub>i</sub> &lt;= col</code></li>
	<li>All the values of <code>cells</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def latestDayToCross(self, row: int, col: int, cells: List[List[int]]) -> int:
        # Disjoint Set (Union-Find)
        class DSU:
            def __init__(self, size):
                self.parent = list(range(size))
                self.rank = [0] * size
                
            def find(self, x):
                if self.parent[x] != x:
                    self.parent[x] = self.find(self.parent[x])
                return self.parent[x]
            
            def union(self, x, y):
                px, py = self.find(x), self.find(y)
                if px != py:
                    if self.rank[px] > self.rank[py]:
                        self.parent[py] = px
                    elif self.rank[px] < self.rank[py]:
                        self.parent[px] = py
                    else:
                        self.parent[py] = px
                        self.rank[px] += 1
        
        # Directions for moving in the matrix
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        # Convert 2D coordinates to 1D index
        def index(r, c):
            return (r - 1) * col + (c - 1)
        
        # Initialize the disjoint set
        dsu = DSU(row * col + 2)
        
        # Top and bottom virtual nodes
        top = row * col
        bottom = row * col + 1
        
        # Binary search on days
        left, right = 0, len(cells) - 1
        result = 0
        
        while left <= right:
            mid = (left + right) // 2
            
            # Reset the disjoint set
            dsu.parent = list(range(row * col + 2))
            dsu.rank = [0] * (row * col + 2)
            
            # Mark cells as water
            grid = [[1] * col for _ in range(row)]
            for i in range(mid + 1):
                r, c = cells[i]
                grid[r - 1][c - 1] = 0
            
            # Connect adjacent land cells
            for r in range(row):
                for c in range(col):
                    if grid[r][c] == 1:
                        for dr, dc in directions:
                            nr, nc = r + dr, c + dc
                            if 0 <= nr < row and 0 <= nc < col and grid[nr][nc] == 1:
                                dsu.union(index(r + 1, c + 1), index(nr + 1, nc + 1))
            
            # Connect top row cells to the top virtual node
            for c in range(col):
                if grid[0][c] == 1:
                    dsu.union(top, index(1, c + 1))
            
            # Connect bottom row cells to the bottom virtual node
            for c in range(col):
                if grid[row - 1][c] == 1:
                    dsu.union(bottom, index(row, c + 1))
            
            # Check if top and bottom are connected
            if dsu.find(top) == dsu.find(bottom):
                result = mid + 1
                left = mid + 1
            else:
                right = mid - 1
        
        return result

# Example usage:
solution = Solution()
row = 2
col = 2
cells = [[1,1],[2,1],[1,2],[2,2]]
print(solution.latestDayToCross(row, col, cells))  # Output: 2
25700