Find Minimum Time to Reach Last Room II - Solution & Explanation

Dijkstra's Algorithm can help find the shortest path in a graph, checking each path's cost in terms of time rather than distance. We treat each room as a node, with edges representing the time it takes to traverse from one node to another (alternating between 1 and 2 seconds). We can use a priority queue to efficiently get the room with the minimum current time cost.

This implementation uses a priority queue to simulate Dijkstra's algorithm. It starts from room (0,0) and explores neighboring rooms with alternating movement times. For each room, it calculates the next possible movement time, checks the constraint with moveTime, and updates with the shortest known time if possible, iterating until it reaches the goal room. If it reaches the end, it returns the current time; otherwise, it continues exploring. The end condition helps in returning the minimum time as updates are based on the earliest time of availability and reachability.

A dynamic programming approach builds a 2D array of size n x m to store the minimum time to reach each room. Starting from (0,0), iterate over each room, calculating the minimum time needed to reach its adjacent rooms with alternating movement times. This approach leverages dynamic properties, updating reachable rooms based on the prior values in the grid plus the movement costs.

This solution uses a 2D dp array to compute the minimum time required to reach each room. Starting from the first room, it iterates through each room (i, j), calculating the minimum time for its adjacent cells based on the current time and alternating increment. The time calculation considers whether an extra second is needed, based on the alternating pattern. It returns the value at the bottom right corner of the dp table.

Using Dijkstra's algorithm is suitable for this problem as it involves finding the shortest path on a weighted graph. Here, the grid can be transformed into a graph where each cell is a node, and edges exist between adjacent cells with weights alternating between 1 and 2 seconds. With moveTime constraints at each cell, we maintain the earliest possible time we can reach each cell, ensuring compliance with these conditions.

This implementation of Dijkstra's algorithm uses a priority queue to explore the shortest paths to each cell efficiently. Every time a new cell is considered, the algorithm checks if moving to that cell is feasible given the constraints in the moveTime matrix. By maintaining a dynamic list via a priority queue, we ensure that we explore the minimum required time paths first, efficiently finding the least time path to the bottom-right corner.

This approach utilizes breadth-first search (BFS), treating move times as conditions while propagating the earliest possible time one can enter each room. Each cell is monitored for the minimum time that satisfies the given moveTime constraint, essentially treating move times as a boundary condition we expand upon.

This solution arranges a BFS traversal alongside a DP table to retain the minimum time for reaching each cell. The traversal leverages a queue to explore neighboring cells in layers based on the constraints provided by moveTime. The key is to maintain updated and valid timings ensuring compliance with the staggered move durations, naturally accommodating the alternating time costs on movements.

We define a two-dimensional array dist, where dist[i][j] represents the minimum time required to reach room (i, j) from the starting point. Initially, we set all elements in the dist array to infinity, and then set the dist value of the starting point (0, 0) to 0.

We use a priority queue pq to store each state, where each state consists of three values (d, i, j), representing the time d required to reach room (i, j) from the starting point. Initially, we add the starting point (0, 0, 0) to pq.

In each iteration, we take the front element (d, i, j) from pq. If (i, j) is the endpoint, we return d. If d is greater than dist[i][j], we skip this state. Otherwise, we enumerate the four adjacent positions (x, y) of (i, j). If (x, y) is within the map, we calculate the final time t from (i, j) to (x, y) as t = max(moveTime[x][y], dist[i][j]) + (i + j) bmod 2 + 1. If t is less than dist[x][y], we update the value of dist[x][y] and add (t, x, y) to pq.

The time complexity is O(n times m times log (n times m)), and the space complexity is O(n times m). Here, n and m are the number of rows and columns of the map, respectively.