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+# 3342. Find Minimum Time to Reach Last Room II
+import heapq
+
+"""
+there is a dungeon with 'n x m' rooms arranged as a grid. you are given a 2d
+array 'moveTime' of size 'n x m' where 'moveTime[i][j]' represents the
+minimum time in seconds when you can start moving to that room. you start
+from the room '(0, 0)' at time 't = 0' and can move to an adjacent room.
+moving between adjacent rooms takes one second for one move and two seconds
+for the next, alternating between the two. return the minimum time to reach
+the room '(n - 1, m - 1)'.
+"""
+
+
+class Solution(object):
+ def minTimeToReach(self, moveTime):
+ """
+ :type moveTime: List[List[int]]
+ :rtype: int
+ """
+ n, m = len(moveTime), len(moveTime[0])
+ dp = [["inf"] * m for i in range(n)]
+ heap = [[0, 0, 0]]
+ moveTime[0][0] = -1
+ while heap:
+ t, i, j = heapq.heappop(heap)
+ if t >= dp[i][j]:
+ continue
+ if i == n - 1 and j == m - 1:
+ return t
+ dp[i][j] = t
+ for di, dj in [[1, 0], [0, 1], [-1, 0], [0, -1]]:
+ x, y = i + di, j + dj
+ c = (i + j) % 2 + 1
+ if 0 <= x < n and 0 <= y < m and dp[x][y] == "inf":
+ heapq.heappush(heap, [max(moveTime[x][y], t) + c, x, y])
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minTimeToReach(moveTime=[[0, 4], [4, 4]]))
+ print(obj.minTimeToReach(moveTime=[[0, 0, 0, 0], [0, 0, 0, 0]]))
+ print(obj.minTimeToReach(moveTime=[[0, 1], [1, 2]]))