<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>

from bisect import bisect_left

class Mono:
    def __init__(self):
        self.I = []
        self.V = []
        self.N = 0
    def put(self, v):
        while self.V and self.V[-1]>=v:
            self.V.pop()
            self.I.pop()
        self.V.append(v)
        self.I.append(self.N)
        self.N += 1
    def query(self, g):
        ii = self.N - g
        j = bisect_left(self.I,ii)
        assert(self.I[j]>=ii)
        return self.V[j]


class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        Inf = 10**9
        M = len(grid)
        N = len(grid[0])
        H = [Mono() for _ in range(M)]
        V = [Mono() for _ in range(N)]
        H[-1].put(1)
        V[-1].put(1)
        for j in range(N-2,-1,-1):
            g = grid[-1][j]
            if g==0:
                v = Inf
            else:
                v = 1+H[-1].query(g)
            H[-1].put(v)
            V[j].put(v)
        for i in range(M-2,-1,-1):
            g = grid[i][-1]
            if g==0:
                v = Inf
            else:    
                v = 1+V[-1].query(g)
            H[i].put(v)
            V[-1].put(v)
            for j in range(N-2,-1,-1):
                g = grid[i][j]
                if g==0:
                    v = Inf
                else:
                    v1 = V[j].query(g)
                    v2 = H[i].query(g)
                    v = 1+min(v1,v2)
                H[i].put(v)
                V[j].put(v)
        r = H[0].query(1)
        self.H = H
        self.V = V
        return r if r<Inf else -1
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