Introduction

In this blog post, we will explore how to solve the LeetCode problem "3342" 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 one second for one move and two seconds for the next, alternating between the two. 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: 7 Explanation: The minimum time required is 7 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 two seconds. Example 2: Input: moveTime = [[0,0,0,0],[0,0,0,0]] Output: 6 Explanation: The minimum time required is 6 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 two seconds. At time t == 3, move from room (1, 1) to room (1, 2) in one second. At time t == 4, move from room (1, 2) to room (1, 3) in two seconds. Example 3: Input: moveTime = [[0,1],[1,2]] Output: 4 Constraints: 2 <= n == moveTime.length <= 750 2 <= m == moveTime[i].length <= 750 0 <= moveTime[i][j] <= 109

Explanation

• Dijkstra's Algorithm: Use Dijkstra's algorithm to find the shortest path from the starting room (0, 0) to the destination room (n-1, m-1). The cost of moving between adjacent rooms alternates between 1 and 2.
  • Time Management: When considering a move to a neighboring room, ensure the current time is at least the moveTime[i][j] of the target room before making the move.
  • Alternating Costs: Keep track of the cost of the last move (1 or 2) and alternate it for the next move.

• 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, 1)]  # (time, row, col, cost) cost is 1 or 2 for the next move

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

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

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

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

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

                if new_time < dist[new_row][new_col]:
                    dist[new_row][new_col] = new_time
                    heapq.heappush(pq, (new_time, new_row, new_col, 3 - cost)) # Alternate the cost

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