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Diffstat (limited to 'find_min_time_reach_last_room1.py')
-rw-r--r-- | find_min_time_reach_last_room1.py | 41 |
1 files changed, 41 insertions, 0 deletions
diff --git a/find_min_time_reach_last_room1.py b/find_min_time_reach_last_room1.py new file mode 100644 index 0000000..3ce1848 --- /dev/null +++ b/find_min_time_reach_last_room1.py @@ -0,0 +1,41 @@ +# 3341. Find Minimum Time to Reach Last Room I +import heapq + +""" +there is a dungeo 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 at '(0, 0)' at time 't = 0' and can move to an anyacent room. +moving between anyacent rooms takes exactly one second. 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)] + h = [[0, 1, 0, 0]] + moveTime[0][0] = -1 + while h: + t, c, i, j = heapq.heappop(h) + if t >= dp[i][j]: + continue + if i == n - 1 and j == m - 1: + return t + dp[i][j] = t + for nx, ny in [[1, 0], [0, 1], [-1, 0], [0, -1]]: + x, y = i + nx, j + ny + if 0 <= x < n and 0 <= y < m and dp[x][y] == "inf": + heapq.heappush(h, [max(moveTime[x][y] + c, t + c), 2 - 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]])) + print(obj.minTimeToReach(moveTime=[[0, 1], [1, 2]])) |