Minimum Time Takes to Reach Destination Without Drowning Python/Java solution

Implement two BFS passes. First compute water arrival times using a queue seeded with all water cells. Then perform BFS from the start cell while checking that the player's time is less than the stored water time. This logic translates cleanly to Python, Java, C++, Go, and TypeScript using standard queue-based BFS.

What is the best approach for Minimum Time Takes to Reach Destination Without Drowning?

The most efficient approach uses two BFS traversals. The first multi-source BFS computes the earliest time water reaches each cell. The second BFS starts from the player's position and only moves into cells where the player arrives strictly before the water. This approach runs in O(m*n) time and O(m*n) space.

Is Minimum Time Takes to Reach Destination Without Drowning asked at Google/Amazon/Meta?

Grid BFS problems with spreading constraints frequently appear in interviews at companies like Amazon, Google, and Meta. Variants include problems where fire spreads, zombies infect cells, or water floods terrain. The core pattern is multi-source BFS combined with shortest path traversal.

What data structure is used in Minimum Time Takes to Reach Destination Without Drowning?

The main data structure is a queue used for breadth-first search. Two BFS traversals are performed: one multi-source queue for water expansion and another queue for the player's movement. A matrix is used to store the earliest water arrival time for each cell.

What is the time complexity of Minimum Time Takes to Reach Destination Without Drowning?

The optimal solution runs in O(m*n) time where m and n are the grid dimensions. One BFS computes water arrival times and another BFS finds the shortest safe path for the player. Each cell is processed at most once per traversal, so the total work is linear in the number of cells.

Description Hints 3 Solution Video 16:28

2814. Minimum Time Takes to Reach Destination Without Drowning

Q: Is Minimum Time Takes to Reach Destination Without Drowning easy or hard?

The problem is rated Hard because it combines multiple BFS concepts. You must model environmental changes over time, compute water spread with multi-source BFS, and then enforce those constraints during pathfinding. Managing timing conditions correctly is the main challenge.

Q: How to solve Minimum Time Takes to Reach Destination Without Drowning in O(m*n)?

Run a multi-source BFS from all water cells to compute when water floods each position. Store these times in a grid. Then run a second BFS from the start cell and move only if the player's arrival time is strictly less than the water arrival time. This ensures every move stays ahead of the flood while exploring the grid only once.

Hard54.7% acceptancePremium elsewhere, free here

You are given an n * m 0-indexed grid of string land. Right now, you are standing at the cell that contains "S", and you want to get to the cell containing "D". There are three other types of cells in this land:

".": These cells are empty.
"X": These cells are stone.
"*": These cells are flooded.

At each second, you can move to a cell that shares a side with your current cell (if it exists). Also, at each second, every empty cell that shares a side with a flooded cell becomes flooded as well.
There are two problems ahead of your journey:

You can't step on stone cells.
You can't step on flooded cells since you will drown (also, you can't step on a cell that will be flooded at the same time as you step on it).

Return the minimum time it takes you to reach the destination in seconds, or -1 if it is impossible.

Note that the destination will never be flooded.

Example 1:

Input: land = [["D",".","*"],[".",".","."],[".","S","."]]
Output: 3
Explanation: The picture below shows the simulation of the land second by second. The blue cells are flooded, and the gray cells are stone.
Picture (0) shows the initial state and picture (3) shows the final state when we reach destination. As you see, it takes us 3 second to reach destination and the answer would be 3.
It can be shown that 3 is the minimum time needed to reach from S to D.

Example 2:

Input: land = [["D","X","*"],[".",".","."],[".",".","S"]]
Output: -1
Explanation: The picture below shows the simulation of the land second by second. The blue cells are flooded, and the gray cells are stone.
Picture (0) shows the initial state. As you see, no matter which paths we choose, we will drown at the 3^rd second. Also the minimum path takes us 4 seconds to reach from S to D.
So the answer would be -1.

Example 3:

Input: land = [["D",".",".",".","*","."],[".","X",".","X",".","."],[".",".",".",".","S","."]]
Output: 6
Explanation: It can be shown that we can reach destination in 6 seconds. Also it can be shown that 6 is the minimum seconds one need to reach from S to D.

Constraints:

2 <= n, m <= 100
land consists only of "S", "D", ".", "*" and "X".
Exactly one of the cells is equal to "S".
Exactly one of the cells is equal to "D".

Topics

Hints open one at a time

Hint 1

Think of using breadth-first search.

Hint 2

Use a BFS to find for each cell the time at which it will become flooded.

Hint 3

Another BFS then simulates your movement, taking into account information gathered by the first BFS.

16:28

Still stuck?Read the written solution, or watch a walkthrough.

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[["D",".","*"],[".",".","."],[".","S","."]]

Input

[["D","X","*"],[".",".","."],[".",".","S"]]

Input

[["D",".",".",".","*","."],[".","X",".","X",".","."],[".",".",".",".","S","."]]

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