<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code>. Your initial position is at the <strong>top-left</strong> cell <code>(0, 0)</code>.</p>

<p>Starting from the cell <code>(i, j)</code>, you can move to one of the following cells:</p>

<ul>
	<li>Cells <code>(i, k)</code> with <code>j &lt; k &lt;= grid[i][j] + j</code> (rightward movement), or</li>
	<li>Cells <code>(k, j)</code> with <code>i &lt; k &lt;= grid[i][j] + i</code> (downward movement).</li>
</ul>

<p>Return <em>the minimum number of cells you need to visit to reach the <strong>bottom-right</strong> cell</em> <code>(m - 1, n - 1)</code>. If there is no valid path, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/25/ex1.png" style="width: 271px; height: 171px;" />
<pre>
<strong>Input:</strong> grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The image above shows one of the paths that visits exactly 4 cells.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/25/ex2.png" style="width: 271px; height: 171px;" />
<pre>
<strong>Input:</strong> grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>The image above shows one of the paths that visits exactly 3 cells.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/26/ex3.png" style="width: 181px; height: 81px;" />
<pre>
<strong>Input:</strong> grid = [[2,1,0],[1,0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be proven that no path exists.
</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;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt; m * n</code></li>
	<li><code>grid[m - 1][n - 1] == 0</code></li>
</ul>

class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        def find_root(x, fa):
            if x == fa[x]:
                return x
            fa[x] = find_root(fa[x], fa)
            return fa[x]

        m, n = len(grid), len(grid[0])
        r_fa = [[j for j in range(n)] for i in range(m)]
        c_fa = [[i for i in range(m)] for j in range(n)]
        visited = [[False for j in range(n)] for i in range(m)]
        
        q = [(0, 0)]
        dis, l, r = 1, 0, 1
        visited[0][0] = True

        while l < r:
            for idx in range(l, r):
                i, j = q[idx]
                if (i == m - 1) and (j == n - 1):
                    return dis
                x = find_root(i, c_fa[j]) + 1
                while (x < m) and (x <= i + grid[i][j]):
                    if not visited[x][j]:
                        visited[x][j] = True
                        q.append((x, j))
                    c_fa[j][x - 1] = find_root(x, c_fa[j])
                    x = c_fa[j][x - 1] + 1
                
                y = find_root(j, r_fa[i]) + 1
                while (y < n) and (y <= j + grid[i][j]):
                    if not visited[i][y]:
                        visited[i][y] = True
                        q.append((i, y))
                    r_fa[i][y - 1] = find_root(y, r_fa[i])
                    y = r_fa[i][y - 1] + 1

            l = r
            r = len(q)
            dis += 1

        return -1

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