Sync LeetCode submission Runtime - 2683 ms (33.90%), Memory - 93.3 MB (54.51%)

Author: jimit105

3628-find-minimum-time-to-reach-last-room-ii/README.md

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+<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>

3628-find-minimum-time-to-reach-last-room-ii/solution.py

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+# Approach: Shortest Path + Dijkstra
+
+# n = no. of rows, m = no. of cols
+# Time: O(mn log (mn))
+# Space: O(mn)
+
+import heapq
+
+class State:
+    def __init__(self, x, y, dis):
+        self.x = x
+        self.y = y
+        self.dis = dis
+
+    def __lt__(self, other):
+        return self.dis < other.dis
+
+
+class Solution:
+    def minTimeToReach(self, moveTime):
+        n = len(moveTime)
+        m = len(moveTime[0])
+        inf = float("inf")
+        d = [[inf] * m for _ in range(n)]
+        v = [[0] * m for _ in range(n)]
+
+        dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
+
+        d[0][0] = 0
+        q = []
+        heapq.heappush(q, State(0, 0, 0))
+
+        while q:
+            s = heapq.heappop(q)
+            if v[s.x][s.y]:
+                continue
+            if s.x == n - 1 and s.y == m - 1:
+                break
+            v[s.x][s.y] = 1
+            for dx, dy in dirs:
+                nx, ny = s.x + dx, s.y + dy
+                if not (0 <= nx < n and 0 <= ny < m):
+                    continue
+                dist = max(d[s.x][s.y], moveTime[nx][ny]) + (s.x + s.y) % 2 + 1
+                if d[nx][ny] > dist:
+                    d[nx][ny] = dist
+                    heapq.heappush(q, State(nx, ny, dist))
+
+        return d[n - 1][m - 1]