<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 sortedcontainers import SortedList
class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        # for each row and each column, keep track of which elements we haven't seen
        # sorted list of indicies
        # then when we consider node (i, j), binary search in rows[i] to find j + 1
        # loop through unseen elements, removing those we can reach
        # same for columns[j]
        # O(mnlogmn)
        m = len(grid)
        n = len(grid[0])
        rows = [SortedList(range(1, n)) for _ in range(m)]
        cols = [SortedList(range(1, m)) for _ in range(n)]


        # outside of that optimization this is just a shortest path problem
        # dequeue contains triples of length, i, j
        dq = collections.deque()
        dq.append((1, 0, 0))
        while len(dq) > 0:
            l, i, j = dq.popleft()
            if i == m - 1 and j == n - 1:
                return l
            toRemove = []
            for v in rows[i].irange(j, j + grid[i][j]):
                dq.append((l + 1, i, v))
                toRemove.append(v)
            for v in toRemove:
                rows[i].remove(v)
            toRemove = []
            for v in cols[j].irange(i, i + grid[i][j]):
                dq.append((l + 1, v, j))
                toRemove.append(v)
            for v in toRemove:
                cols[j].remove(v)
           
        return -1
            
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