<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:
        m, n = len(grid), len(grid[0])
        mx = m + n # use it as the limit, there is no more than m + n - 1 steps if the solution exists
        h = [[] for i in range(m)] # horizontal monotonic stacks
        v = [[] for i in range(n)] # vertical monotonic stacks 
        
        for r in range(m - 1, -1, -1):
            for c in range(n - 1, -1, -1):
                if r == m - 1 and c == n - 1: # init stacks
                    h[r].append((c, 1)) # (column index, number of steps)
                    v[c].append((r, 1)) # (row index, number of steps)
                else:
                    s, mn = grid[r][c], mx # s - allowed steps from the current cell, mn - minimum possible steps
                    while len(h[r]) > 1 and h[r][-2][0] - s <= c:
                        h[r].pop() # we can jump to a further cell with less steps to the target 
                    mn = min(mn, h[r][-1][1] + 1 if h[r] and h[r][-1][0] - s <= c else mx)
                    while len(v[c]) > 1 and v[c][-2][0] - s <= r:
                        v[c].pop() # we can jump to a further cell with less steps to the target 
                    mn = min(mn, v[c][-1][1] + 1 if v[c] and v[c][-1][0] - s <= r else mx)
                    while h[r] and h[r][-1][1] >= mn:
                        h[r].pop() # clear less optimal horizontal solutions, if it's better to move vertically
                    h[r].append((c, mn))
                    while v[c] and v[c][-1][1] >= mn:
                        v[c].pop() # clear less optimal vertical solutions, if it's better to move horizontally
                    v[c].append((r, mn))

        mn = h[0][-1][1]
        return mn if mn < mx else -1
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