Introduction

In this blog post, we will explore how to solve the LeetCode problem "3341" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is a dungeon 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 (0, 0) at time t = 0 and can move to an adjacent room. Moving between adjacent rooms takes exactly one second. Return the minimum time to reach the room (n - 1, m - 1). Two rooms are adjacent if they share a common wall, either horizontally or vertically. Example 1: Input: moveTime = [[0,4],[4,4]] Output: 6 Explanation: The minimum time required is 6 seconds. At time t == 4, move from room (0, 0) to room (1, 0) in one second. At time t == 5, move from room (1, 0) to room (1, 1) in one second. Example 2: Input: moveTime = [[0,0,0],[0,0,0]] Output: 3 Explanation: The minimum time required is 3 seconds. At time t == 0, move from room (0, 0) to room (1, 0) in one second. At time t == 1, move from room (1, 0) to room (1, 1) in one second. At time t == 2, move from room (1, 1) to room (1, 2) in one second. Example 3: Input: moveTime = [[0,1],[1,2]] Output: 3 Constraints: 2 <= n == moveTime.length <= 50 2 <= m == moveTime[i].length <= 50 0 <= moveTime[i][j] <= 109

Explanation

Here's a solution to the problem, addressing efficiency and optimality:

• Dijkstra's Algorithm: Use Dijkstra's algorithm to find the shortest path from the starting cell (0, 0) to the destination cell (n-1, m-1). Dijkstra's is suitable for finding shortest paths in a graph with non-negative edge weights.
• Time Management: At each cell, calculate the minimum arrival time by considering the moveTime[i][j] constraint. If the current time is less than moveTime[i][j], wait until moveTime[i][j] before moving to the next adjacent cell.

• Priority Queue: Use a priority queue (heap) to efficiently select the cell with the minimum arrival time to explore next.

• Runtime Complexity: O(n m log(n m)), Storage Complexity: O(n m)

Code

    import heapq

def minTimeToReachRoom(moveTime):
    n = len(moveTime)
    m = len(moveTime[0])

    dist = [[float('inf')] * m for _ in range(n)]
    dist[0][0] = 0

    pq = [(0, 0, 0)]  # (time, row, col)

    while pq:
        time, row, col = heapq.heappop(pq)

        if time > dist[row][col]:
            continue

        if row == n - 1 and col == m - 1:
            return time

        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for dr, dc in directions:
            new_row = row + dr
            new_col = col + dc

            if 0 <= new_row < n and 0 <= new_col < m:
                wait_time = max(0, moveTime[new_row][new_col] - time)
                new_time = time + wait_time + 1

                if new_time < dist[new_row][new_col]:
                    dist[new_row][new_col] = new_time
                    heapq.heappush(pq, (new_time, new_row, new_col))

    return dist[n-1][m-1]