Maximum Vacation Days

Hard

Datadog

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ArraysDynamic Programming

You are given a country consisting of n cities. The cities are labeled from 0 to n - 1.

You are also given a 0-indexed n x n integer matrix flights representing the ability to fly between cities. If flights[i][j] == 1, it means that you can fly from city i to city j.

You are also given a 0-indexed integer matrix days of size n x k representing available vacation days in each city. You will stay in a city for one week (7 days) to take the vacation days, and you can only stay in one city each week. You can fly to another city at the end of the week:

You will start at city 0 on week 0.
You can only fly to city j at the end of week i if flights[currentCity][j] == 1.
You can stay in a city for multiple weeks.

Return the maximum number of vacation days you can take if you are given k weeks.

Example 1:

Input: flights = [[0,1,1],[1,0,0],[0,0,0]], days = [[0,2,1],[0,0,0],[0,0,0]]
Output: 3
Explanation:
Week 0 : Stay at city 0, you can take 0 vacation days, 
Week 1 : Fly to city 1, you can take 0 vacation days,
Week 2 : Fly to city 0, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

Week 0 : Stay at city 0, you can take 0 vacation days,
Week 1 : Fly to city 2, you can take 0 vacation days,
Week 2 : Stay at city 2, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

Week 0 : Stay at city 0, you can take 0 vacation days,
Week 1 : Stay at city 0, you can take 0 vacation days,
Week 2 : Fly to city 1, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

It is impossible to get more than 3 vacation days.

Example 2:

Input: flights = [[0,0,0],[0,0,0],[0,0,0]], days = [[1,1,1],[7,7,7],[7,7,7]]
Output: 21
Explanation:
Stay at city 0 for all 3 weeks to get 1 + 1 + 1 = 3 vacation days.
Stay at city 1 for all 3 weeks to get 7 + 7 + 7 = 21 vacation days.
Stay at city 2 for all 3 weeks to get 7 + 7 + 7 = 21 vacation days.
It is impossible to get more than 21 vacation days.

Constraints:

n == flights.length
n == flights[i].length
n == days.length
k == days[i].length
1 <= n, k <= 100
flights[i][j] is either 0 or 1.
days[i][j] is in the range [0, 1000].

Try coding it in LeetCode

Maximum Vacation Days

Hard

Datadog

5 views

Topics:

ArraysDynamic Programming

You are given a country consisting of n cities. The cities are labeled from 0 to n - 1.

You are also given a 0-indexed n x n integer matrix flights representing the ability to fly between cities. If flights[i][j] == 1, it means that you can fly from city i to city j.

You are also given a 0-indexed integer matrix days of size n x k representing available vacation days in each city. You will stay in a city for one week (7 days) to take the vacation days, and you can only stay in one city each week. You can fly to another city at the end of the week:

You will start at city 0 on week 0.
You can only fly to city j at the end of week i if flights[currentCity][j] == 1.
You can stay in a city for multiple weeks.

Return the maximum number of vacation days you can take if you are given k weeks.

Example 1:

Input: flights = [[0,1,1],[1,0,0],[0,0,0]], days = [[0,2,1],[0,0,0],[0,0,0]]
Output: 3
Explanation:
Week 0 : Stay at city 0, you can take 0 vacation days, 
Week 1 : Fly to city 1, you can take 0 vacation days,
Week 2 : Fly to city 0, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

Week 0 : Stay at city 0, you can take 0 vacation days,
Week 1 : Fly to city 2, you can take 0 vacation days,
Week 2 : Stay at city 2, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

Week 0 : Stay at city 0, you can take 0 vacation days,
Week 1 : Stay at city 0, you can take 0 vacation days,
Week 2 : Fly to city 1, you can take 0 vacation days,
Total vacation days = 0 + 0 + 0 = 0.

It is impossible to get more than 3 vacation days.

Example 2:

Input: flights = [[0,0,0],[0,0,0],[0,0,0]], days = [[1,1,1],[7,7,7],[7,7,7]]
Output: 21
Explanation:
Stay at city 0 for all 3 weeks to get 1 + 1 + 1 = 3 vacation days.
Stay at city 1 for all 3 weeks to get 7 + 7 + 7 = 21 vacation days.
Stay at city 2 for all 3 weeks to get 7 + 7 + 7 = 21 vacation days.
It is impossible to get more than 21 vacation days.

Constraints:

n == flights.length
n == flights[i].length
n == days.length
k == days[i].length
1 <= n, k <= 100
flights[i][j] is either 0 or 1.
days[i][j] is in the range [0, 1000].

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:

Can you provide the constraints for the number of cities (`n`), the number of weeks (`k`), and the number of vacation days allowed per city per week?
Are the flight schedules one-way or round-trip? That is, if I can fly from city A to city B in week X, can I necessarily fly from city B to city A in week X?
If it's impossible to take any vacation (due to travel restrictions or no vacation days offered), what should the function return?
If there are multiple paths that result in the same maximum number of vacation days, is any one of those paths acceptable, or is there a tie-breaking rule?
Is it possible for the `days` array to contain negative numbers, or are the number of vacation days always non-negative?

Brute Force Solution

Approach

The brute force method in this scenario involves exploring every single path to determine the one that yields the highest number of vacation days. We will systematically check all possible travel schedules to find the best one. This means considering every possible city to visit at each week of vacation.

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

Start at week zero in your home city.
For each week, consider flying to every possible city you are allowed to fly to based on the flight schedules.
Calculate the vacation days you would get if you chose to stay in that city for that week.
Repeat this process for every possible city you could fly to each week, taking into account the flight schedules and vacation days.
Continue this process until you have considered every possible sequence of cities for all weeks.
Keep track of the total vacation days for each sequence of cities you considered.
Finally, choose the city sequence that results in the highest number of total vacation days.

Code Implementation

def max_vacation_days_brute_force(
    flight_schedules, vacation_days_per_city
): 

    number_of_cities = len(vacation_days_per_city)
    number_of_weeks = len(vacation_days_per_city[0])

    maximum_vacation_days = 0

    def calculate_vacation_days(
        current_week_index, current_city_index, current_vacation_days
    ): 
        nonlocal maximum_vacation_days

        # Base case: Reached the end of the vacation
        if current_week_index == number_of_weeks:
            maximum_vacation_days = max(
                maximum_vacation_days, current_vacation_days
            )
            return

        # Iterate through all possible cities to fly to
        for next_city_index in range(number_of_cities): 
            # Must fly to next city or stay in current city
            if (
                flight_schedules[current_city_index][next_city_index] == 1
            ): 
                vacation_days_in_next_city = vacation_days_per_city[
                    next_city_index
                ][current_week_index]
                calculate_vacation_days(
                    current_week_index + 1,
                    next_city_index,
                    current_vacation_days + vacation_days_in_next_city,
                )

    # Start recursion from city 0 during week 0
    calculate_vacation_days(
        0, 0, vacation_days_per_city[0][0]
    ) 

    return maximum_vacation_days

Big(O) Analysis

Time Complexity

O(N^K) – Let N be the number of cities and K be the number of weeks. The brute force approach explores every possible path of cities to visit for each week. In the worst case, for each of the K weeks, we could potentially visit any of the N cities. This results in a branching factor of N at each week, and we explore this for K weeks. Therefore, the time complexity is approximately N multiplied by itself K times, which equals O(N^K).

Space Complexity

O(N^W) – The brute force approach explores every possible path through the cities. For each week, we consider all possible cities to fly to. In the worst-case scenario, we are building up a recursion call stack, or implicitly storing the sequence of cities visited, up to the number of weeks. If there are N cities and W weeks, this branching exploration leads to a space complexity proportional to the number of possible paths we have explored and need to keep track of to compute vacation days, hence N^W where N is the number of cities and W is the number of weeks.

Optimal Solution

Approach

We want to find the maximum vacation days possible by strategically choosing which city to visit each week. The optimal approach uses a concept similar to finding the shortest path to a destination. We build up the solution week by week, always keeping track of the best possible number of vacation days we can achieve up to that week.

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

Start by figuring out how many vacation days we can get if we start our vacation in each city on the first week.
For each subsequent week, look at each city and figure out the best way to get there based on where we were the week before. We can only travel to cities that are reachable from our previous city.
When calculating the total vacation days for each city in a given week, consider the maximum vacation days we could have accumulated by the previous week in any reachable city, plus the vacation days offered by the current city.
Keep track of the best possible total vacation days we can have at the end of each week for each city.
After processing all the weeks, the answer is the maximum number of vacation days among all the cities for the last week.

Code Implementation

def max_vacation_days(flights, vacation_days):
    number_of_cities = len(flights)
    number_of_weeks = len(vacation_days[0])

    # Initialize the maximum vacation days for each city for the first week.
    maximum_vacation_days_for_week = [0] * number_of_cities
    for city in range(number_of_cities):
        if flights[0][city] == 1:
            maximum_vacation_days_for_week[city] = vacation_days[city][0]

    # Iterate over the weeks to find the max vacation days for each city.
    for week in range(1, number_of_weeks):
        previous_vacation_days = maximum_vacation_days_for_week[:]
        maximum_vacation_days_for_week = [0] * number_of_cities

        for current_city in range(number_of_cities):
            # Iterate through possible previous cities.
            for previous_city in range(number_of_cities):
                # Check if we can reach the current city from the previous city.
                if flights[previous_city][current_city] == 1:
                    # Update the max vacation days for the current city.
                    maximum_vacation_days_for_week[current_city] = max(
                        maximum_vacation_days_for_week[current_city],
                        previous_vacation_days[previous_city] + vacation_days[current_city][week]
                    )

            #If starting in city 0, can always stay put
            maximum_vacation_days_for_week[current_city] = max(
                maximum_vacation_days_for_week[current_city],
                maximum_vacation_days_for_week[current_city] + vacation_days[current_city][week] if week == 1 and flights[0][current_city]==0 else maximum_vacation_days_for_week[current_city]
            )

    # Find the maximum vacation days among all cities in the last week.
    max_vacation = 0
    for vacation_days_in_city in maximum_vacation_days_for_week:
        max_vacation = max(max_vacation, vacation_days_in_city)

    return max_vacation

Big(O) Analysis

Time Complexity

O(n * m * n) – The algorithm iterates through each week (m weeks). For each week, it iterates through each city (n cities). Inside the inner loop, it iterates through each possible city from the previous week (n cities) to check reachability. The nested loops cause n * m * n operations. Thus, the time complexity is O(n * m * n), where n is the number of cities and m is the number of weeks. We can simplify this to O(n² * m).

Space Complexity

O(N) – The algorithm maintains a table to store the maximum vacation days achievable up to each week for each city. This table, often implemented as a 2D array, has dimensions W x N, where W is the number of weeks and N is the number of cities. Therefore, the auxiliary space required is proportional to the number of cities, N, for each week. Since we only need to keep track of the vacation days for the current week, we can overwrite values from the previous weeks; only two rows (current and previous) are ever needed. Hence, the space complexity is O(N) because we store maximum vacation days for each city for the current and previous weeks, but not for all weeks.

Edge Cases

Case	How to Handle
Empty 'days' or 'flights' arrays	Return 0 if either 'days' or 'flights' is empty, as no vacation is possible.
'days' array with different lengths for each city	Throw an IllegalArgumentException or return an error code, since this violates the problem constraints
City can't reach any other city (isolated node)	The DP transition should account for unreachable cities by initializing the maximum vacation days to negative infinity if the travel isn't possible.
City can only reach itself	Ensure the initial state and DP transitions allow for staying in the same city and accumulating vacation days.
Very large 'days' or 'flights' arrays leading to potential memory issues	Consider using a space-optimized DP approach, such as rolling arrays, to reduce memory footprint.
Negative values in 'days' array (invalid vacation days)	Throw an IllegalArgumentException or return an error code as the number of days should be non-negative.
Cycles in the 'flights' array where it becomes impossible to visit a useful city again within the remaining weeks.	The DP approach should implicitly handle cycles by considering all possible paths and selecting the one with maximum vacation days.
Integer overflow when calculating maximum vacation days	Use a larger data type (long) for storing the maximum vacation days to prevent potential overflow issues during calculations.