How do you solve Find Minimum Time to Reach Last Room II on LeetCode?

Find Minimum Time to Reach Last Room II (LeetCode #3342) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding the alternating move times is crucial for calculating the total time.

What is the brute force solution for Find Minimum Time to Reach Last Room II?

The brute force approach for Find Minimum Time to Reach Last Room II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths from the starting room to the destination room. It calculates the time taken for each path, considering the alternating move times, and returns the minimum time found. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find Minimum Time to Reach Last Room II?

Find Minimum Time to Reach Last Room II uses the following concepts: Array, Graph Theory, Heap (Priority Queue), Matrix, Shortest Path. The recommended patterns to study are: Dijkstra's Algorithm, Graph Traversal, Priority Queue.

What are the interview tips for Find Minimum Time to Reach Last Room II?

Explain your thought process clearly while coding. Consider edge cases, such as minimum grid sizes. Practice explaining your approach to someone else before the interview.

What are common mistakes when solving Find Minimum Time to Reach Last Room II?

Not considering the alternating move times correctly. Failing to handle visited states, leading to cycles.

#3342

Find Minimum Time to Reach Last Room II

Medium

ArrayGraph TheoryHeap (Priority Queue)MatrixShortest PathDijkstra's AlgorithmGraph TraversalPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a priority queue (min-heap) to implement Dijkstra's algorithm, which efficiently finds the shortest path in a weighted grid. Each state considers the current room, the time taken to reach it, and whether the last move was odd or even.

⚙️

Algorithm

3 steps

1Step 1: Initialize a priority queue with the starting room (0, 0) and the initial time.
2Step 2: While the queue is not empty, extract the room with the minimum time.
3Step 3: For each adjacent room, calculate the new time based on whether the last move was odd or even, and push the new state into the queue if it improves the time.

solution.py19 lines

1import heapq
2
3def minTime(moveTime):
4    n, m = len(moveTime), len(moveTime[0])
5    pq = [(moveTime[0][0], 0, 0, True)]  # (time, x, y, odd)
6    visited = set()
7    while pq:
8        time, x, y, odd = heapq.heappop(pq)
9        if (x, y) == (n - 1, m - 1):
10            return time
11        if (x, y, odd) in visited:
12            continue
13        visited.add((x, y, odd))
14        for dx, dy in [(1, 0), (0, 1), (-1, 0), (0, -1)]:
15            nx, ny = x + dx, y + dy
16            if 0 <= nx < n and 0 <= ny < m:
17                next_time = time + (1 if odd else 2) + moveTime[nx][ny]
18                heapq.heappush(pq, (next_time, nx, ny, not odd))
19    return -1

ℹ

Complexity note: The time complexity is O(n log n) because we are using a priority queue to process each room efficiently. The space complexity is O(n) due to the storage of visited states and the priority queue.

1Understanding the alternating move times is crucial for calculating the total time.
2Using a priority queue allows us to efficiently find the shortest path.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.