Find Minimum Time to Reach Last Room I

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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 after which the room opens and can be moved to. 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

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 constraints on the size of the input array `rooms`? Can the array be empty?
What are the possible values for the waiting time at each room (`wait_time[i]`)? Can they be negative or zero?
Is the `wait_time` array guaranteed to have the same length as the `rooms` array?
If it's impossible to reach the last room, what should I return? Should I throw an exception, or return a specific value like -1 or `Infinity`?
Could you provide an example of a scenario where the last room cannot be reached, and what the expected output should be in that case?

Brute Force Solution

Approach

The brute force approach is like trying every possible path to the last room. We'll explore each potential route, step-by-step, and see how long it takes to get to the end.

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

Start at the first room and consider all possible rooms we can go to next.
For each of these next rooms, calculate the total time it would take to reach them from the start.
From each of those rooms, consider all possible rooms we can go to *next*, and again calculate the total time to reach these new rooms.
Continue exploring all possible paths, always calculating the total time to reach each room on each path.
Eventually, you'll reach the last room through many different paths. Calculate the total time for each of these paths.
Compare the total times for all paths that lead to the last room.
The smallest total time among all paths is the minimum time to reach the last room.

Code Implementation

def find_minimum_time_to_reach_last_room_brute_force(time_taken):
    number_of_rooms = len(time_taken)
    minimum_total_time = float('inf')

    def explore_paths(current_room, current_time):
        nonlocal minimum_total_time

        # If we've reached the last room,
        # update the minimum time if necessary
        if current_room == number_of_rooms - 1:
            minimum_total_time = min(minimum_total_time, current_time)
            return

        # Explore all possible next rooms
        for next_room in range(number_of_rooms):
            # Avoid cycles by skipping the current room
            if next_room != current_room:

                time_to_next_room = time_taken[current_room][next_room]

                # Recursively explore the path
                explore_paths(next_room, current_time + time_to_next_room)

    # Begin exploring paths from the first room at time 0
    explore_paths(0, 0)

    # Return the minimum time found or -1 if no path exists
    if minimum_total_time == float('inf'):
        return -1
    else:
        return minimum_total_time

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach explores all possible paths to the last room. In the worst case, the algorithm might visit all possible permutations of the rooms. For n rooms, there are n! (n factorial) possible paths. Therefore, the time complexity of this approach is O(n!).

Space Complexity

O(N^N) – The brute force approach described explores all possible paths, implying a branching factor roughly proportional to N (the number of rooms) at each step. Since the maximum path length can also be proportional to N (visiting each room once), the number of paths explored can grow up to N^N in the worst case. Storing the total time for each of these paths to reach the last room requires memory proportional to the number of paths. Thus, the space complexity is O(N^N).

Optimal Solution

Approach

To efficiently determine the minimum time, we will focus on scheduling visits to each room in a way that minimizes waiting. The key insight is to visit rooms in a specific order based on when they become available, which avoids unnecessary delays.

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

First, calculate the time you can actually enter each room; this depends on the room's designated wait time.
Next, sort the rooms based on their 'earliest entry time', meaning the smallest waiting time comes first.
Then, start visiting rooms in this sorted order, keeping track of the current time.
For each room, check if the current time is earlier than its 'earliest entry time'. If it is, you'll have to wait until that time to enter the room, updating your 'current time' accordingly.
Add one unit of time to your 'current time' to account for the time spent visiting the room.
Continue this process for each room in the sorted order.
The final 'current time' after visiting the last room is the minimum time needed to reach the last room.

Code Implementation

def find_minimum_time_to_reach_last_room_i(wait_times):
    number_of_rooms = len(wait_times)
    entry_times = []

    for room_index, wait_time in enumerate(wait_times):
        entry_times.append((room_index, wait_time))

    # Sort rooms by earliest entry time.
    entry_times.sort(key=lambda x: x[1])

    current_time = 0

    for room_index, wait_time in entry_times:
        # Need to wait if current time is before the earliest entry time.
        if current_time < wait_time:
            current_time = wait_time

        current_time += 1

    return current_time

Big(O) Analysis

Time Complexity

O(n log n) – The provided solution involves calculating the earliest entry time for each of the n rooms, which takes O(n) time. The dominant operation is sorting the rooms based on their earliest entry time. Efficient sorting algorithms like merge sort or quicksort typically have a time complexity of O(n log n). Visiting each room in the sorted order then takes O(n) time. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The provided algorithm creates a sorted list of rooms based on their 'earliest entry time'. This sorted list requires auxiliary space proportional to the number of rooms, which is N. The algorithm does not use recursion or any other data structures that scale with the input size beyond this sorted list. Therefore, the space complexity is dominated by the storage of the sorted rooms.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there are no rooms to traverse.
Array with only one room (length 1)	Return 0, as the starting room is also the last room.
Large input array (nearing memory limits)	Use an efficient data structure like a priority queue that scales well with large inputs to avoid memory issues.
All rooms have the same waiting time	The algorithm should still proceed linearly, accumulating the waiting time at each room.
Waiting times are very large (potential integer overflow)	Use long data type to prevent potential overflow during cumulative time calculation.
Rooms are arranged in decreasing order of waiting time, causing frequent waits	The algorithm must correctly handle this worst-case scenario, potentially leading to longer execution time but still producing the correct result.
Zero waiting time for all rooms	The algorithm should efficiently traverse the array without adding any waiting time.
Input array is already sorted in ascending order based on room index	Algorithm should work correctly on this input, adding wait times when necessary.

Null or empty input array

How to Handle:

Return 0 immediately as there are no rooms to traverse.

Array with only one room (length 1)

How to Handle:

Return 0, as the starting room is also the last room.

Large input array (nearing memory limits)

How to Handle:

Use an efficient data structure like a priority queue that scales well with large inputs to avoid memory issues.

All rooms have the same waiting time

How to Handle:

The algorithm should still proceed linearly, accumulating the waiting time at each room.

Waiting times are very large (potential integer overflow)

How to Handle:

Use long data type to prevent potential overflow during cumulative time calculation.

Rooms are arranged in decreasing order of waiting time, causing frequent waits

How to Handle:

The algorithm must correctly handle this worst-case scenario, potentially leading to longer execution time but still producing the correct result.

Zero waiting time for all rooms

How to Handle:

The algorithm should efficiently traverse the array without adding any waiting time.

Input array is already sorted in ascending order based on room index

How to Handle:

Algorithm should work correctly on this input, adding wait times when necessary.