<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
from typing import List

class MonotonicQueue:
    """A monotonic queue that maintains values in increasing order."""
    def __init__(self):
        self.indices = []  # Stores indices of elements
        self.values = []   # Stores values of elements
        self.count = 0     # Total number of elements processed

    def put(self, value):
        """Add a value to the queue, maintaining monotonicity."""
        # Remove all values greater than or equal to the new value
        while self.values and self.values[-1] >= value:
            self.values.pop()
            self.indices.pop()
        self.values.append(value)
        self.indices.append(self.count)
        self.count += 1

    def query(self, gap):
        """Query the smallest value that is at least 'gap' positions away."""
        target_index = self.count - gap
        j = bisect_left(self.indices, target_index)
        assert self.indices[j] >= target_index
        return self.values[j]


class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        INF = 10**9
        M, N = len(grid), len(grid[0])
        
        # Initialize monotonic queues for each row and column
        horizontal_queues = [MonotonicQueue() for _ in range(M)]
        vertical_queues = [MonotonicQueue() for _ in range(N)]

        # Initialize the bottom-right cell
        horizontal_queues[-1].put(1)
        vertical_queues[-1].put(1)

        # Process the bottom row
        for j in range(N-2, -1, -1):
            jump = grid[-1][j]
            value = 1 + horizontal_queues[-1].query(jump) if jump > 0 else INF
            horizontal_queues[-1].put(value)
            vertical_queues[j].put(value)

        # Process the rightmost column
        for i in range(M-2, -1, -1):
            jump = grid[i][-1]
            value = 1 + vertical_queues[-1].query(jump) if jump > 0 else INF
            horizontal_queues[i].put(value)
            vertical_queues[-1].put(value)

        # Process the rest of the grid
        for i in range(M-2, -1, -1):
            for j in range(N-2, -1, -1):
                jump = grid[i][j]
                if jump == 0:
                    value = INF
                else:
                    vertical_min = vertical_queues[j].query(jump)
                    horizontal_min = horizontal_queues[i].query(jump)
                    value = 1 + min(vertical_min, horizontal_min)
                horizontal_queues[i].put(value)
                vertical_queues[j].put(value)

        # Query the result for the top-left cell
        result = horizontal_queues[0].query(1)
        return result if result < INF else -1
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