<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 typing import List
from collections import deque

class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])  # Dimensions of the grid
        
        # Initialize the BFS queue with the starting cell (0,0) and its cost
        queue = deque([(0, 0, 1)])
        
        # Set to keep track of visited cells
        visited = set([(0, 0)])
        
        while queue:
            lvsize = len(queue)  # Number of cells at the current level
            
            # Process all cells at the current level
            for _ in range(lvsize):
                i, j, cost = queue.popleft()
                
                # If we've reached the bottom-right cell, return the cost
                if i == m-1 and j == n-1:
                    return cost
                
                # Check downward movements
                start = i + 1
                while start < m and start < i + grid[i][j] + 1:
                    if (start, j) in visited:
                        break  # Stop if we hit a visited cell
                    queue.append((start, j, cost+1))
                    visited.add((start, j))
                    start += 1
                
                # Check upward movements (optimization)
                end = min(i + grid[i][j], m-1)
                while end >= 0 and end > i:
                    if (end, j) in visited:
                        break  # Stop if we hit a visited cell
                    queue.append((end, j, cost+1))
                    visited.add((end, j))
                    end -= 1
                
                # Check rightward movements
                start = j + 1
                while start < n and start < j + grid[i][j] + 1:
                    if (i, start) in visited:
                        break  # Stop if we hit a visited cell
                    queue.append((i, start, cost+1))
                    visited.add((i, start))
                    start += 1
                
                # Check leftward movements (optimization)
                end = min(j + grid[i][j], n-1)
                while end >= 0 and end > j:
                    if (i, end) in visited:
                        break  # Stop if we hit a visited cell
                    queue.append((i, end, cost+1))
                    visited.add((i, end))
                    end -= 1
        
        # If we've exhausted all possible moves without reaching the target
        return -1
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