<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>

INF = 10 ** 9
from heapq import *


class Solution:
    def minimumVisitedCells(self, a: List[List[int]]) -> int:
        
        n = len(a)
        m = len(a[0])
                
        dp_ij = 0
        
        n_pq = [[] for _ in range(n)]
        m_pq = [[] for _ in range(m)]
        
        heappush(n_pq[0], (dp_ij, a[0][0]))
        heappush(m_pq[0], (dp_ij, a[0][0]))
                
            
        for i in range(n):
            for j in range(m):
                
                if i == 0 and j == 0:
                    continue

                while m_pq[j] and i > m_pq[j][0][1]:
                    heappop(m_pq[j])
                    
                while n_pq[i] and j > n_pq[i][0][1]:
                    heappop(n_pq[i])
                
                case_1 = m_pq[j][0][0] if m_pq[j] else INF
                case_2 = n_pq[i][0][0] if n_pq[i] else INF
                                
                
                dp_ij = 1 + min(case_1, case_2)
                
                heappush(m_pq[j], (dp_ij, a[i][j] + i))
                heappush(n_pq[i], (dp_ij, a[i][j] + j))
                
        
        return dp_ij + 1 if dp_ij < INF else -1
63100