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Diffstat (limited to 'min_time_visit_cell_grid.py')
-rw-r--r-- | min_time_visit_cell_grid.py | 44 |
1 files changed, 44 insertions, 0 deletions
diff --git a/min_time_visit_cell_grid.py b/min_time_visit_cell_grid.py new file mode 100644 index 0000000..aab6075 --- /dev/null +++ b/min_time_visit_cell_grid.py @@ -0,0 +1,44 @@ +# 2577. Minimum Time to Visit a Cell In a Grid +import heapq + +""" +you are given an 'm x n' matrix 'grid' consisting of non-negative integers +where 'grid[row][col]' represents the minimum time required to be able to +visit the cell '(row, col)', which means you can visit the cell '(row, col)' +only when the time you visit it is greater than or equal to the value. you +are standing in the top left cell of the matrix in the 0th second and you +must move to any adjacent cell in the four directions: up, down, left, and +right. each move you make takes 1 second. return the minimum time required in +which you can visit the bottom right cell of the matrix. if you cannot visit +the bottomright cell, then return -1. +""" + + +class Solution(object): + def minimumTime(self, grid): + """ + :type grid: List[List[int]] + :rtype: int + """ + if grid[0][1] > 1 and grid[1][0] > 1: + return -1 + m, n = len(grid), len(grid[0]) + vis, pq = set(), [(grid[0][0], 0, 0)] + while pq: + time, row, col = heapq.heappop(pq) + if row == m - 1 and col == n - 1: + return time + if (row, col) in vis: + continue + vis.add((row, col)) + for dr, dc in [(0, 1), (0, -1), (1, 0), (-1, 0)]: + r, c = row + dr, col + dc + if 0 <= r < m and 0 <= c < n and (r, c) not in vis: + wait = 1 if ((grid[r][c] - time) % 2 == 0) else 0 + heapq.heappush(pq, (max(time + 1, grid[r][c] + wait), r, c)) + + +if __name__ == "__main__": + obj = Solution() + print(obj.minimumTime(grid=[[0, 1, 3, 2], [5, 1, 2, 5], [4, 3, 8, 6]])) + print(obj.minimumTime(grid=[[0, 2, 4], [3, 2, 1], [1, 0, 4]])) |