<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 MinStack:
    def __init__(self):
        self.stk = []
    def query(self, pos):
        K = len(self.stk)
        left, right = 0, K-1
        if left > right:
            return 10**6
        while left < right:
            mid = left + (right - left) // 2
            if self.stk[mid][0] > pos:
                left = mid + 1
            else:
                right = mid
        if self.stk[left][0] > pos:
            left += 1
        if left == K:
            return 10**6
        return self.stk[left][1]

    def update(self, pos, val):
        while self.stk and self.stk[-1][1] >= val:
            self.stk.pop()
        self.stk.append([pos, val])
class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        M, N = len(grid), len(grid[0])
        dp = [[(10**6) for _ in range(N)] for _ in range(M)]
        rowStk = [MinStack() for _ in range(M)]
        colStk = [MinStack() for _ in range(N)]
        for i in range(M-1, -1, -1):
            for j in range(N-1, -1, -1):
                if i == M-1 and j == N-1:
                    dp[i][j] = 1
                else:
                    dp[i][j] = min(dp[i][j], rowStk[i].query(min(N-1,grid[i][j]+j)))
                    dp[i][j] = min(dp[i][j], colStk[j].query(min(M-1, grid[i][j]+i)))
                    if dp[i][j] < (10**6):
                        dp[i][j] += 1
                rowStk[i].update(j, dp[i][j])
                colStk[j].update(i, dp[i][j])
        return dp[0][0] if dp[0][0] < (10**6) else -1
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