<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:
        INF = float('inf');
        dp = [[INF for _ in range(len(grid[0]))] for _ in range(len(grid))];
        col_heaps = [[] for _ in range(len(grid[0]))];
        row_heaps = [[] for _ in range(len(grid))];

        for row in range(len(grid)):
            for col in range(len(grid[0])):
                colHeap = col_heaps[col]
                rowHeap = row_heaps[row]
                
                # base case: starting from the beginning square is one action
                if col == row == 0:
                    heapq.heappush(colHeap, (1, 0, 0))
                    heapq.heappush(rowHeap, (1, 0, 0))
                    dp[0][0] = 1
                    continue

                bestCol = INF
                while(colHeap):
                    best1, row1, col1 = colHeap[0]
                    delta = grid[row1][col1]

                    # can we reach current square?
                    if delta + row1 >= row:
                        bestCol = best1
                        break

                    # if not, toss him
                    heapq.heappop(colHeap)

                bestRow = INF
                while(rowHeap):
                    best2, row2, col2 = rowHeap[0]
                    delta = grid[row2][col2]

                    # can we reach current square?
                    if delta + col2 >= col:
                        bestRow = best2
                        break

                    # if not, toss him
                    heapq.heappop(rowHeap)
                    
                rating = min(bestRow, bestCol)
                dp[row][col] = rating + 1

                heapq.heappush(colHeap, (rating + 1, row, col))
                heapq.heappush(rowHeap, (rating + 1, row, col))
        
        return (dp[-1][-1]) if dp[-1][-1] != INF else -1
71500