Problem Description

You are in a dungeon structured as a grid consisting of n rows and m columns. You must navigate through this grid to reach your destination. The grid is represented by a 2D array moveTime, where each cell moveTime[i][j] specifies the earliest possible time in seconds that you can start moving into that room (i, j). You begin your journey from the top-left corner of the grid, room (0, 0), at time t = 0. Each move from one room to an adjacent room (horizontally or vertically) takes exactly one second. Your objective is to find the minimum time necessary to reach the bottom-right corner of the grid, room (n - 1, m - 1).

Intuition

To solve this problem, we can utilize Dijkstra's algorithm, which is well-suited for finding the shortest path in graphs with non-negative weights. Here, the grid can be visualized as a weighted graph where each cell is a node and each move from one cell to an adjacent cell represents an edge with a weight of one second.

1. Set Up the Distance Array: Initialize a 2D array dist to keep track of the minimum time to reach each room. Set the initial room’s (0, 0) time to 0 and all others to infinity (inf), as we're interested in the shortest time.

2. Priority Queue for Exploration: Use a priority queue to explore the rooms. We start by pushing the starting room (0, 0) with a time of 0 into the priority queue.

3. Explore Using Dijkstra's Approach: With the priority queue, iteratively expand the least-time path available. For the current room (i, j), look at its adjacent rooms. Calculate the time it would take to get to an adjacent room, which is the maximum of the moveTime requirement and the accumulated time so far plus one second for the move.

4. Update Times and Continue: If moving to the adjacent room (x, y) can be done in a time earlier than what’s currently recorded in dist[x][y], update it and push this new state into the priority queue.

5. Terminate on Goal Room: When reaching the target room (n-1, m-1) with the least time, return this time as it's the shortest time needed given the constraints of the moveTime grid.

This intuitive approach leverages both the spatial nature of grid movement and the temporal constraints provided, efficiently finding the optimal route in terms of time.

Learn more about Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The problem of finding the minimum time to navigate the dungeon grid while adhering to the constraints of the moveTime array can be efficiently addressed using Dijkstra's Algorithm. This algorithm is effective in determining the shortest path in a graph, especially when dealing with non-negative edge weights, as is the case here where each move takes one second.

Step-by-Step Implementation:

1. Initialize the Distance Array:

   • Create a 2D array dist of the same dimensions as moveTime which will hold the minimum time required to reach each room. Initially, set every room's time to infinity (inf), except for the starting room (0, 0), which is set to 0 since you start there at time 0.

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

2. Set Up the Priority Queue:

   • Use a priority queue to facilitate the exploration of the least-time path first. This priority queue will store tuples in the format (d, i, j), where d is the time to reach the room (i, j). Begin by inserting the starting room with its initial time into the queue.

   pq = [(0, 0, 0)]

3. Dijkstra's Exploration:

   • While the priority queue is not empty, extract the room (i, j) with the current minimum time d.
   • If this room is the destination (n - 1, m - 1), return d as the shortest time.
   • Skip this room if the current time d exceeds dist[i][j], as this indicates a previously discovered shorter path.
   • For each of the four possible adjacent rooms (up, down, left, right), calculate the potential time t to move there, which is the maximum of the move's time constraint and the accumulated time plus one second.

4. Update and Push New States:

   • If the new calculated time t to an adjacent room (x, y) is less than the currently recorded time in dist[x][y], update dist[x][y] and push the new state (t, x, y) into the priority queue for further exploration.

   for a, b in pairwise(dirs):
       x, y = i + a, j + b
       if 0 <= x < n and 0 <= y < m:
           t = max(moveTime[x][y], dist[i][j]) + 1
           if dist[x][y] > t:
               dist[x][y] = t
               heappush(pq, (t, x, y))

5. Termination:

   • The algorithm naturally terminates as soon as the destination is dequeued from the priority queue with the least possible time path due to Dijkstra's greedy nature, ensuring optimality.

This structured approach efficiently navigates the grid using the properties of the priority queue to always process the most promising path next, thus ensuring that the solution is optimal in terms of the time constraints imposed.

Example Walkthrough

Let's illustrate the solution with a small example. Assume the moveTime grid is as follows:

moveTime = [
    [0, 2, 2],
    [2, 3, 5],
    [5, 5, 8]
]

This grid has 3 rows and 3 columns. We need to determine the minimum time needed to move from the top-left corner (0, 0) to the bottom-right corner (2, 2).

Step-by-Step Walkthrough:

1. Initialize the Distance Array:

   We initialize dist as:

   dist = [
       [0, inf, inf],
       [inf, inf, inf],
       [inf, inf, inf]
   ]

   The starting room (0, 0) is set to 0.

2. Set Up the Priority Queue:

   Start with a priority queue containing the starting point:

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

3. Dijkstra's Exploration:

   Process rooms by expanding from the least-time path:

   • Step 1: Dequeue (0, 0, 0) from the queue.

     • Explore adjacent rooms (0, 1) and (1, 0).
     • To (0, 1): Time = max(2, 0) + 1 = 3. Update dist[0][1] = 3 and add (3, 0, 1) to the queue.
     • To (1, 0): Time = max(2, 0) + 1 = 3. Update dist[1][0] = 3 and add (3, 1, 0) to the queue.

   • Step 2: Dequeue (3, 0, 1) (because 3 is the smallest).

     • Explore (0, 2) and (1, 1).
     • To (0, 2): Time = max(2, 3) + 1 = 4. Update dist[0][2] = 4 and add (4, 0, 2) to the queue.
     • To (1, 1): Time = max(3, 3) + 1 = 4. Update dist[1][1] = 4 and add (4, 1, 1) to the queue.

   • Step 3: Dequeue (3, 1, 0) (next smallest time).

     • Explore (2, 0) and (1, 1).
     • To (2, 0): Time = max(5, 3) + 1 = 6. Update dist[2][0] = 6 and add (6, 2, 0) to the queue.

   • Step 4: Dequeue (4, 0, 2).

     • Explore (1, 2).
     • To (1, 2): Time = max(5, 4) + 1 = 6. Update dist[1][2] = 6 and add (6, 1, 2) to the queue.

4. Continue Exploration:

   • Follow a similar process until you reach (2, 2):
     • From (4, 1, 1), explore to (2, 1), and so on.
     • Eventually, reach (2, 2) with time 7.

5. Termination:

   When (2, 2) is reached, the queue provides the minimum time 8. Thus, the minimum time required to navigate from (0, 0) to (2, 2) respecting the moveTime constraints is 8.

This walkthrough showcases the step-by-step application of Dijkstra's algorithm in traversing the grid based on the constraints provided by moveTime.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n × m × log(n × m)). This arises from the fact that the algorithm uses Dijkstra's algorithm with a priority queue (min-heap). Pushing and popping from the heap costs O(log k) where k is the number of elements in the heap, and k can reach up to n × m in the worst case, where n is the number of rows and m is the number of columns. For each cell, the algorithm considers up to four neighboring cells, leading to O(n × m × log(n × m)) in total.

The space complexity of the code is O(n × m). This is due to the storage of the dist array that keeps track of the minimum time to reach each cell, and the priority queue pq that can store up to n × m elements.

Learn more about how to find time and space complexity quickly.