Find Minimum Time to Reach Last Room II
<p>There is a dungeon with <code>n x m</code> rooms arranged as a grid.</p>

<p>You are given a 2D array <code>moveTime</code> of size <code>n x m</code>, where <code>moveTime[i][j]</code> represents the <strong>minimum</strong> time in seconds when you can <strong>start moving</strong> to that room. You start from the room <code>(0, 0)</code> at time <code>t = 0</code> and can move to an <strong>adjacent</strong> room. Moving between <strong>adjacent</strong> rooms takes one second for one move and two seconds for the next, <strong>alternating</strong> between the two.</p>

<p>Return the <strong>minimum</strong> time to reach the room <code>(n - 1, m - 1)</code>.</p>

<p>Two rooms are <strong>adjacent</strong> if they share a common wall, either <em>horizontally</em> or <em>vertically</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">moveTime = [[0,4],[4,4]]</span></p>

<p><strong>Output:</strong> 7</p>

<p><strong>Explanation:</strong></p>

<p>The minimum time required is 7 seconds.</p>

<ul>
	<li>At time <code>t == 4</code>, move from room <code>(0, 0)</code> to room <code>(1, 0)</code> in one second.</li>
	<li>At time <code>t == 5</code>, move from room <code>(1, 0)</code> to room <code>(1, 1)</code> in two seconds.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">moveTime = [[0,0,0,0],[0,0,0,0]]</span></p>

<p><strong>Output:</strong> 6</p>

<p><strong>Explanation:</strong></p>

<p>The minimum time required is 6 seconds.</p>

<ul>
	<li>At time <code>t == 0</code>, move from room <code>(0, 0)</code> to room <code>(1, 0)</code> in one second.</li>
	<li>At time <code>t == 1</code>, move from room <code>(1, 0)</code> to room <code>(1, 1)</code> in two seconds.</li>
	<li>At time <code>t == 3</code>, move from room <code>(1, 1)</code> to room <code>(1, 2)</code> in one second.</li>
	<li>At time <code>t == 4</code>, move from room <code>(1, 2)</code> to room <code>(1, 3)</code> in two seconds.</li>
</ul>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">moveTime = [[0,1],[1,2]]</span></p>

<p><strong>Output:</strong> 4</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n == moveTime.length &lt;= 750</code></li>
	<li><code>2 &lt;= m == moveTime[i].length &lt;= 750</code></li>
	<li><code>0 &lt;= moveTime[i][j] &lt;= 10<sup>9</sup></code></li>
</ul>
Graph; Array; Matrix; Shortest Path; Heap (Priority Queue)
class Solution:
    def minTimeToReach(self, moveTime: List[List[int]]) -> int:
        n, m = len(moveTime), len(moveTime[0])
        dist = [[inf] * m for _ in range(n)]
        dist[0][0] = 0
        pq = [(0, 0, 0)]
        dirs = (-1, 0, 1, 0, -1)
        while 1:
            d, i, j = heappop(pq)
            if i == n - 1 and j == m - 1:
                return d
            if d > dist[i][j]:
                continue
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < m:
                    t = max(moveTime[x][y], dist[i][j]) + (i + j) % 2 + 1
                    if dist[x][y] > t:
                        dist[x][y] = t
                        heappush(pq, (t, x, y))
