Find Minimum Time to Reach Last Room II

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GraphsDynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the `times` array, and what is the maximum number of rooms I can expect?
Is it possible to have `wait` times be negative, zero, or non-integer?
If it's impossible to reach the last room, what should I return?
If there are multiple paths that lead to the last room in the minimum time, is any one of them acceptable, or is there a preference?
Is the `times` array guaranteed to be non-empty?

Brute Force Solution

Approach

We want to find the fastest way to get to the last room. The brute force approach explores every single possible path we could take, trying them all out to see which one is the quickest. This involves simulating every possible order of visiting rooms.

Here's how the algorithm would work step-by-step:

Start at the first room.
Consider every possible next room to visit.
For each of these possible next rooms, consider every possible room to visit after that.
Keep doing this until you have created every single possible route from the first room to the last room.
For each route, calculate the total time it takes to travel that route.
Compare the total times of all the routes.
Choose the route with the smallest total time; that's the fastest way.

Code Implementation

def find_minimum_time_brute_force(time_matrix):
    number_of_rooms = len(time_matrix)
    visited_rooms = [False] * number_of_rooms
    best_time = float('inf')

    def solve(current_room, current_time, rooms_visited_count):
        nonlocal best_time

        # We've reached the last room, check if the time is the best
        if current_room == number_of_rooms - 1:
            best_time = min(best_time, current_time)
            return

        # If all rooms have been visited except the last one, no path exists
        if rooms_visited_count == number_of_rooms - 1 and current_room != number_of_rooms - 1:
            return

        visited_rooms[current_room] = True

        # Explore every possible next room
        for next_room in range(number_of_rooms):
            if not visited_rooms[next_room]:

                # Recursive call to explore the next room
                solve(next_room, current_time + time_matrix[current_room][next_room], rooms_visited_count + 1)

        visited_rooms[current_room] = False

    # Begin the search from the first room
    solve(0, 0, 1)

    # If no path exists, return -1
    if best_time == float('inf'):
        return -1
    return best_time

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers every possible permutation of visiting the n rooms. For n rooms, there are n! (n factorial) possible orderings. For each permutation, we calculate the total time to traverse the path, which takes O(n) time to sum the waiting times and travel times. Therefore, the overall time complexity is dominated by generating and evaluating all permutations. This leads to a time complexity of O(n * n!), which is simplified to O(n!).

Space Complexity

O(N!) – The brute force approach explores all possible paths, which means generating all permutations of the rooms to visit (excluding the starting room since it's fixed). Storing each permutation requires O(N) space, where N is the number of rooms. Since there are N! (N factorial) possible permutations, the algorithm needs to store up to N! paths, each of length roughly N, leading to significant memory usage. The auxiliary space required to hold these permutations and track visited rooms grows factorially with the number of rooms. Therefore, the space complexity is O(N!).

Optimal Solution

Approach

The key to solving this problem efficiently is to work backwards and use information about the last rooms to determine the earliest possible arrival time at previous rooms. We utilize a priority queue to always process rooms with the currently known earliest arrival time.

Here's how the algorithm would work step-by-step:

Start from the last room because we know the final arrival time is already set at 0. The ultimate goal is to work backward from the last room to find the minimum time to reach the first room.
Maintain a collection of room arrival times, and add the last room with time 0 to the collection to kick things off.
Keep track of two things: current arrival time at the previous room and total waiting time required at a room, based on the provided array.
Iterate through the rooms backwards. For each room, use the waiting time to determine the total time needed to reach the current room.
The total arrival time becomes the earliest time we could have reached the room before the current room.
Put this total arrival time into the collection and then go to the room before that one.
Whenever you put a room arrival time into the collection, make sure you don't already have a better arrival time for that room. If you do, replace it.
By repeatedly working backward and always choosing the minimum possible arrival time for each room, you eventually determine the minimum possible arrival time for the first room.

Code Implementation

import heapq

def find_minimum_time_to_reach_last_room_ii(waiting_times):
    number_of_rooms = len(waiting_times)
    arrival_times = [float('inf')] * number_of_rooms
    arrival_times[number_of_rooms - 1] = 0

    priority_queue = [(0, number_of_rooms - 1)]

    while priority_queue:
        current_arrival_time, current_room = heapq.heappop(priority_queue)

        # Skip processing if we've already found a faster route to this room
        if current_arrival_time > arrival_times[current_room]:
            continue

        if current_room == 0:
            return current_arrival_time

        previous_room = current_room - 1

        # Critical step: calculate arrival time at the previous room
        time_to_reach_current_room = current_arrival_time + waiting_times[previous_room]

        arrival_time_previous_room = time_to_reach_current_room + 1

        # Update arrival time if a better path is found
        if arrival_time_previous_room < arrival_times[previous_room]:
            arrival_times[previous_room] = arrival_time_previous_room
            heapq.heappush(priority_queue, (arrival_time_previous_room, previous_room))

    return -1

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates backward through the n rooms. Inside the loop, a priority queue (or similar data structure like a heap) is used to store and retrieve the earliest arrival times for rooms. The insertion and extraction operations on a priority queue of size n take O(log n) time. Since the priority queue operations are performed within the loop that iterates n times, the overall time complexity becomes O(n log n).

Space Complexity

O(N) – The algorithm maintains a collection (likely a priority queue or similar structure) of room arrival times, potentially storing an arrival time for each of the N rooms. It also keeps track of the earliest arrival time for each room which could require storing up to N values. Therefore, the auxiliary space required scales linearly with the number of rooms, N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty wait times array	Return -1 or throw an exception to indicate invalid input.
Number of rooms is zero or negative	Return -1 or throw an exception to indicate invalid input since you can't have a negative or zero number of rooms.
Wait times array contains negative values	Treat them as invalid and return -1 or throw exception, assuming wait times cannot be negative.
Wait times array contains extremely large values that may lead to integer overflow	Use long data type to prevent overflow or perform modulo operations if applicable to keep values within range.
First room has a very long wait time pushing the time very far out	The algorithm should correctly propagate the long wait time forward influencing the minimum time to reach the last room.
Loop or cycle in the room dependencies (if the problem implicitly suggests dependencies)	The problem should specify whether such dependencies can exist, and if so the algorithm must detect the cycle and terminate or return an appropriate error indicator if a valid solution is impossible.
Maximum size of number of rooms and large wait times causing time to reach last room to exceed maximum integer size.	Use long data type and check for overflow during calculations, and return a large value or throw exception if overflow is detected.
All wait times are zero	The algorithm should correctly calculate the minimum time as the number of rooms minus one since each room can be entered immediately after entering the prior room.

Null or empty wait times array

How to Handle:

Return -1 or throw an exception to indicate invalid input.

Number of rooms is zero or negative

How to Handle:

Return -1 or throw an exception to indicate invalid input since you can't have a negative or zero number of rooms.

Wait times array contains negative values

How to Handle:

Treat them as invalid and return -1 or throw exception, assuming wait times cannot be negative.

Wait times array contains extremely large values that may lead to integer overflow

How to Handle:

Use long data type to prevent overflow or perform modulo operations if applicable to keep values within range.

First room has a very long wait time pushing the time very far out

How to Handle:

The algorithm should correctly propagate the long wait time forward influencing the minimum time to reach the last room.

Loop or cycle in the room dependencies (if the problem implicitly suggests dependencies)

How to Handle:

The problem should specify whether such dependencies can exist, and if so the algorithm must detect the cycle and terminate or return an appropriate error indicator if a valid solution is impossible.

Maximum size of number of rooms and large wait times causing time to reach last room to exceed maximum integer size.

How to Handle:

Use long data type and check for overflow during calculations, and return a large value or throw exception if overflow is detected.

All wait times are zero

How to Handle:

The algorithm should correctly calculate the minimum time as the number of rooms minus one since each room can be entered immediately after entering the prior room.