<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:
        # Fix i,j, to each (i,j), we either go from above or from left
        # Hence, the idea is to record the min steps to reach prrvious cells
        # then update the current one
        # In order to efficiently find the min reachable previous cells,
        # we use heap data structure
        m = len(grid)
        n = len(grid[0])
        if m==1 and n==1:
            return 1
        from_above = [[] for _ in range(n)]
        from_left = [[] for _ in range(m)]
 
        for i in range(m):
            for j in range(n):
                score = float('inf')

                while from_left[i] and from_left[i][0][1] + grid[i][from_left[i][0][1]] < j:
                    heapq.heappop(from_left[i])
                if from_left[i]:
                    score = min(score,from_left[i][0][0])

                while from_above[j] and from_above[j][0][1] + grid[from_above[j][0][1]][j] < i:
                    heapq.heappop(from_above[j])
                if from_above[j]:
                    score = min(score,from_above[j][0][0])

                if i==0 and j==0:
                    heapq.heappush(from_above[0], (1,0))
                    heapq.heappush(from_left[0], (1,0))
                    
                if (i,j)==(m-1,n-1):
                    return score+1 if score!=float('inf') else -1
                if score!=float('inf'):
                    heapq.heappush(from_above[j], (score+1,i))
                    heapq.heappush(from_left[i], (score+1,j))
            



                


        
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