<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:
        n, mm = map(len, (grid, grid[0]))
        hs = [[] for _ in range(n)]
        vs = [[] for _ in range(mm)]
        dp = [[inf] * mm for _ in range(n)]
        hs[- 1].append(mm - 1)
        vs[- 1].append(n - 1)
        dp[- 1][- 1] = 1
        for i in range(n - 1, - 1, - 1):
            for j in range(mm - 1 - int(i + 1 == n), - 1, - 1):
                if grid[i][j] == 0:
                    continue
                cur = inf
                mi, mj = i + grid[i][j], j + grid[i][j]
                l, r = 0, len(hs[i]) - 1
                while l < r:
                    m = (l + r) // 2
                    if hs[i][m] > mj:
                        l = m + 1
                    else:
                        r = m
                if l < len(hs[i]) and hs[i][l] <= mj:
                    cur = min(cur, dp[i][hs[i][l]])
                l, r = 0, len(vs[j]) - 1
                while l < r:
                    m = (l + r) // 2
                    if vs[j][m] > mi:
                        l = m + 1
                    else:
                        r = m
                if l < len(vs[j]) and vs[j][l] <= mi:
                    cur = min(cur, dp[vs[j][l]][j])
                dp[i][j] = cur + 1

                while hs[i] and dp[i][hs[i][- 1]] >= dp[i][j]:
                    hs[i].pop()
                hs[i].append(j)
                while vs[j] and dp[vs[j][- 1]][j] >= dp[i][j]:
                    vs[j].pop()
                vs[j].append(i)

        return dp[0][0] if dp[0][0] != inf else - 1

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