Leetcode #2814: Minimum Time Takes to Reach Destination Without Drowning
In this guide, we solve Leetcode #2814 Minimum Time Takes to Reach Destination Without Drowning in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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".
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Breadth-First Search, Array, Matrix
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
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.
Python Solution
class Solution:
def minimumSeconds(self, land: List[List[str]]) -> int:
m, n = len(land), len(land[0])
vis = [[False] * n for _ in range(m)]
g = [[inf] * n for _ in range(m)]
q = deque()
si = sj = 0
for i, row in enumerate(land):
for j, c in enumerate(row):
match c:
case "*":
q.append((i, j))
case "S":
si, sj = i, j
dirs = (-1, 0, 1, 0, -1)
t = 0
while q:
for _ in range(len(q)):
i, j = q.popleft()
g[i][j] = t
for a, b in pairwise(dirs):
x, y = i + a, j + b
if (
0 <= x < m
and 0 <= y < n
and not vis[x][y]
and land[x][y] in ".S"
):
vis[x][y] = True
q.append((x, y))
t += 1
t = 0
q = deque([(si, sj)])
vis = [[False] * n for _ in range(m)]
vis[si][sj] = True
while q:
for _ in range(len(q)):
i, j = q.popleft()
if land[i][j] == "D":
return t
for a, b in pairwise(dirs):
x, y = i + a, j + b
if (
0 <= x < m
and 0 <= y < n
and g[x][y] > t + 1
and not vis[x][y]
and land[x][y] in ".D"
):
vis[x][y] = True
q.append((x, y))
t += 1
return -1
Complexity
The time complexity is and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.