2409. Count Days Spent Together

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Problem Description

Alice and Bob are both visiting Rome on separate business trips, and we're given specific dates for when they will arrive and when they will leave. The objective is to calculate the total number of days both Alice and Bob will be in Rome simultaneously. We are provided with four dates: arriveAlice and leaveAlice for Alice's visit, and arriveBob and leaveBob for Bob's visit. Each date is given as a string in the "MM-DD" format, where "MM" represents the month and "DD" represents the day.

The problem specifies that the time period we're considering occurs within the same year, and importantly, it is not a leap year, so February has 28 days. The number of days per month can be represented as [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] corresponding to January through December, respectively.

We need to return the number of days that Alice and Bob will be in the city at the same time. This is inclusive of both their arrival and departure dates if they overlap at all.

Intuition

The intuition behind the provided solution is to find the common range of dates when both Alice and Bob are in Rome and then calculate the length of this overlap. The approach includes the following steps:

  1. Determine the latest arrival date between Alice and Bob. This is done by comparing arriveAlice and arriveBob and taking the maximum of the two. This marks the start of the period when both are in Rome.

  2. Determine the earliest leave date between Alice and Bob by comparing leaveAlice and leaveBob and taking the minimum of the two. This marks the end of the period when both are in Rome.

  3. With these boundaries (start and end), calculate the total number of days they overlap. This requires converting the dates into a cumulative day count from the start of the year and then finding the difference between the start and end dates.

  4. The day count for each date is found by summing the total number of days in the months leading up to the given month using the provided array for the days in each month, then adding the day of the month from the date.

  5. Once we have the day counts for both the start and end of the overlap, we calculate the difference. If the latest arrival is after the earliest departure, this would result in a negative number, indicating no overlap. Since we cannot have negative days of overlap, we use the max function to return zero in such cases.

  6. If there is an overlap, we add 1 to the difference since both the start and end dates are inclusive.

By using these steps, we efficiently find the number of days two intervals overlap, which corresponds to the number of days Alice and Bob are in Rome simultaneously.

Learn more about Math patterns.

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Solution Approach

The implementation of the solution uses straightforward calculations and built-in Python functions to compute the number of overlapping days. Below are the specific steps detailed with the algorithms and data structures used:

  1. Determine Overlap Start and End:

    • The solution defines two variables, a and b, using the built-in max and min functions to calculate the latest arrival date and the earliest departure date, respectively.
    • a and b are used to represent the start and end of the overlapping period.

  2. Prepare Data Structure for Days per Month:

    • A tuple named days is used to store the number of days in each month for a non-leap year. This is a fixed-size, immutable data structure, which is appropriate for the constant values representing the days in each month.

  3. Convert Dates to Day Counts:

    • To convert the dates in "MM-DD" format to a cumulative count of days from the start of the year, the solution uses string slicing and the sum function.
    • For example, sum(days[: int(a[:2]) - 1]) calculates the total days up to the start of the month represented by a. It slices the days tuple up to the month index (subtracting 1 since Python uses 0-indexing) and sums the values.
    • Then, it adds the day of the month from a (or b for the end date): int(a[3:]).

  4. Calculate Overlapping Days:

    • With the day counts for the start (x) and end (y) of the overlap determined, the difference y - x gives the number of days in between these dates, not including the start date.
    • Since the problem states the dates are inclusive, 1 is added to the difference.
    • To ensure there is no negative count of overlap (which would represent no overlap at all), the max function is used to return either the calculated overlap or 0 if the overlap calculation is negative: max(y - x + 1, 0).

By working through these steps, the solution effectively transforms the date range comparison problem into one of simple arithmetic and takes advantage of Python's powerful built-in functions and data structures to do so efficiently.

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What's the output of running the following function using the following tree as input? 

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Assume we have the following travel dates for Alice and Bob:

  • Alice's travel dates: arriveAlice = "02-08" and leaveAlice = "02-15"
  • Bob's travel dates: arriveBob = "02-09" and leaveBob = "02-18"

Following the solution approach:

  1. Determine Overlap Start and End:

    • Compare Alice's and Bob's arrival dates, find the latest arrival date: arriveBob is "02-09", which is later than arriveAlice.
    • Compare Alice's and Bob's leave dates, find the earliest leave date: leaveAlice is "02-15", which is earlier than leaveBob.
    • Thus, the overlap starts on "02-09" and ends on "02-15".

  2. Prepare Data Structure for Days per Month:

    • We already have a tuple that represents the days in each month: (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31).

  3. Convert Dates to Day Counts:

    • To calculate the day count for "02-09" (arriveBob), we sum the days for January and add the days in February up to the 9th: 31 + 9 = 40.
    • To calculate the day count for "02-15" (leaveAlice), we sum the days for January and add the days in February up to the 15th: 31 + 15 = 46.

  4. Calculate Overlapping Days:

    • The difference between the day counts is 46 - 40 = 6. Since both the start and end dates are inclusive, we add 1 to this difference, giving us 6 + 1 = 7.
    • The overlap is 7 days, meaning Alice and Bob will be in Rome simultaneously for a period of 7 days.

By executing these steps with the given dates and applying the logic and calculations described in the solution approach, we find that Alice and Bob will be in Rome at the same time for a total of 7 days.

Solution Implementation

1class Solution:
2    def countDaysTogether(self, arrive_alice: str, leave_alice: str, arrive_bob: str, leave_bob: str) -> int:
3        # Retrieve the later arrival date between Alice and Bob
4        later_arrival = max(arrive_alice, arrive_bob)
5      
6        # Retrieve the earlier departure date between Alice and Bob
7        earlier_departure = min(leave_alice, leave_bob)
8      
9        # Tuple containing the number of days in each month
10        days_in_months = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
11      
12        # Calculate the day of the year for the later arrival date
13        arrival_day_of_year = sum(days_in_months[:int(later_arrival[:2]) - 1]) + int(later_arrival[3:])
14      
15        # Calculate the day of the year for the earlier departure date
16        departure_day_of_year = sum(days_in_months[:int(earlier_departure[:2]) - 1]) + int(earlier_departure[3:])
17      
18        # Calculate and return the number of shared days between Alice and Bob
19        # Add 1 because if they arrive and leave on the same day, they have spent 1 day together
20        shared_days = max(departure_day_of_year - arrival_day_of_year + 1, 0)
21        return shared_days
22
1class Solution {
2  
3    // Array to store the number of days in each month considering a non-leap year.
4    private int[] daysInMonth = new int[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
5
6    /**
7     * Calculates the number of days Alice and Bob spend together.
8     *
9     * @param arriveAlice Arrival date of Alice in "MM-DD" format.
10     * @param leaveAlice  Departure date of Alice in "MM-DD" format.
11     * @param arriveBob   Arrival date of Bob in "MM-DD" format.
12     * @param leaveBob    Departure date of Bob in "MM-DD" format.
13     * @return The number of days they spend together.
14     */
15    public int countDaysTogether(
16        String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {
17      
18        // Determine the later arrival date of the two.
19        String laterArrival = arriveAlice.compareTo(arriveBob) < 0 ? arriveBob : arriveAlice;
20        // Determine the earlier leave date of the two.
21        String earlierLeave = leaveAlice.compareTo(leaveBob) < 0 ? leaveAlice : leaveBob;
22      
23        // Convert the dates to the day of the year.
24        int startDay = convertToDateInYear(laterArrival);
25        int endDay = convertToDateInYear(earlierLeave);
26      
27        // Calculate the total number of days they spend together.
28        // Ensure it does not result in a negative number if there is no overlap.
29        return Math.max(endDay - startDay + 1, 0);
30    }
31
32    /**
33     * Converts a date string "MM-DD" to the day of the year.
34     *
35     * @param date String formatted as "MM-DD".
36     * @return The day of the year.
37     */
38    private int convertToDateInYear(String date) {
39        // Parse the month from the date string and adjust for 0-based index.
40        int monthIndex = Integer.parseInt(date.substring(0, 2)) - 1;
41        int dayOfYear = 0;
42      
43        // Add the number of days in the months preceding the given month.
44        for (int i = 0; i < monthIndex; ++i) {
45            dayOfYear += daysInMonth[i];
46        }
47      
48        // Add the day of the month to the total.
49        dayOfYear += Integer.parseInt(date.substring(3));
50        return dayOfYear;
51    }
52}
53
1#include <vector>
2#include <string>
3
4class Solution {
5public:
6    std::vector<int> daysInMonth = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
7
8    // Calculates the number of days Alice and Bob spend together based on their arrival and departure dates
9    int countDaysTogether(std::string arriveAlice, std::string leaveAlice, std::string arriveBob, std::string leaveBob) {
10        // Find the later of the two arrival dates
11        std::string latestArrival = arriveAlice < arriveBob ? arriveBob : arriveAlice;
12        // Find the earlier of the two departure dates
13        std::string earliestDeparture = leaveAlice < leaveBob ? leaveAlice : leaveBob;
14        // Convert dates to day of the year
15        int dayOfYearLatestArrival = convertDateToDayOfYear(latestArrival);
16        int dayOfYearEarliestDeparture = convertDateToDayOfYear(earliestDeparture);
17        // Calculate the number of days they spend together and ensure it's not negative
18        return std::max(0, dayOfYearEarliestDeparture - dayOfYearLatestArrival + 1);
19    }
20
21    // Converts a date in MM-DD format to a number representing the day of the year
22    private:
23    int convertDateToDayOfYear(std::string date) {
24        int month, day;
25        // Parse the month and day from the date string
26        sscanf(date.c_str(), "%d-%d", &month, &day);
27        int dayOfYear = 0;
28        // Accumulate the total number of days from the beginning of the year to the end of the previous month
29        for (int i = 0; i < month - 1; ++i) {
30            dayOfYear += daysInMonth[i];
31        }
32        // Add the days in the current month
33        dayOfYear += day;
34        return dayOfYear;
35    }
36};
37
1// An array storing the number of days in each month, assuming it is not a leap year
2const daysInMonth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
3
4/**
5 * Calculates the number of days Alice and Bob spend together based on their arrival and departure dates
6 * 
7 * @param arriveAlice - The arrival date for Alice in MM-DD format
8 * @param leaveAlice - The departure date for Alice in MM-DD format
9 * @param arriveBob - The arrival date for Bob in MM-DD format
10 * @param leaveBob - The departure date for Bob in MM-DD format
11 * @returns The number of days Alice and Bob spend together
12 */
13function countDaysTogether(arriveAlice: string, leaveAlice: string, arriveBob: string, leaveBob: string): number {
14    // Find the later of the two arrival dates
15    const latestArrival = arriveAlice < arriveBob ? arriveBob : arriveAlice;
16    // Find the earlier of the two departure dates
17    const earliestDeparture = leaveAlice < leaveBob ? leaveAlice : leaveBob;
18    // Convert dates to day of the year
19    const dayOfYearLatestArrival = convertDateToDayOfYear(latestArrival);
20    const dayOfYearEarliestDeparture = convertDateToDayOfYear(earliestDeparture);
21    // Calculate the number of days they spend together and ensure it's not negative
22    return Math.max(0, dayOfYearEarliestDeparture - dayOfYearLatestArrival + 1);
23}
24
25/**
26 * Converts a date in MM-DD format to a number representing the day of the year
27 * 
28 * @param date - The date in MM-DD format
29 * @returns The day of the year based on the date provided
30 */
31function convertDateToDayOfYear(date: string): number {
32    // Parse the month and day from the date string
33    const [month, day] = date.split('-').map(Number);
34    let dayOfYear = 0;
35    // Accumulate the total number of days from the beginning of the year to the end of the previous month
36    for (let i = 0; i < month - 1; ++i) {
37        dayOfYear += daysInMonth[i];
38    }
39    // Add the days in the current month
40    dayOfYear += day;
41    return dayOfYear;
42}
43
44// Example usage:
45// console.log(countDaysTogether('08-15', '08-18', '08-16', '08-20'));
46
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Which of the following is the prefix sum of array [1, 2, 3, 4, 5]?

Time and Space Complexity

The given Python code calculates the number of overlapping days that Alice and Bob spend together based on their arrival and departure dates.

Time Complexity

The time complexity of this code can be considered to be O(1). Here is the breakdown of operations:

  • Two max and min operations on dates: These operations take constant time since the operations are performed on fixed-length strings (date formats).
  • Slicing and accessing elements of the date strings: These operations are also constant in time since dates are fixed-length strings.
  • Two sum operations: The days tuple is a fixed size (12 elements representing the days in each month), so summing over a slice of this tuple is also a constant-time operation.
  • Integer conversion and arithmetic operations are again done in constant time.

Since all the operations above are done in constant time and do not depend on the size of the input, the overall time complexity is O(1).

Space Complexity

The space complexity of the function can also be considered O(1) because:

  • The tuple days has a fixed size of 12.
  • There are only a fixed number of integer and string variables, a, b, x, and y.
  • No additional data structures that grow with the input size are being used.

Therefore, the amount of memory used by the program is constant, irrespective of the input, leading to a space complexity of O(1).

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

Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using input 56?

1KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13    def dfs(path, res):
14        if len(path) == len(digits):
15            res.append(''.join(path))
16            return
17
18        next_number = digits[len(path)]
19        for letter in KEYBOARD[next_number]:
20            path.append(letter)
21            dfs(path, res)
22            path.pop()
23
24    res = []
25    dfs([], res)
26    return res
27
1private static final Map<Character, char[]> KEYBOARD = Map.of(
2    '2', "abc".toCharArray(),
3    '3', "def".toCharArray(),
4    '4', "ghi".toCharArray(),
5    '5', "jkl".toCharArray(),
6    '6', "mno".toCharArray(),
7    '7', "pqrs".toCharArray(),
8    '8', "tuv".toCharArray(),
9    '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13    List<String> res = new ArrayList<>();
14    dfs(new StringBuilder(), res, digits.toCharArray());
15    return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19    if (path.length() == digits.length) {
20        res.add(path.toString());
21        return;
22    }
23    char next_digit = digits[path.length()];
24    for (char letter : KEYBOARD.get(next_digit)) {
25        path.append(letter);
26        dfs(path, res, digits);
27        path.deleteCharAt(path.length() - 1);
28    }
29}
30
1const KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13    let res = [];
14    dfs(digits, [], res);
15    return res;
16}
17
18function dfs(digits, path, res) {
19    if (path.length === digits.length) {
20        res.push(path.join(''));
21        return;
22    }
23    let next_number = digits.charAt(path.length);
24    for (let letter of KEYBOARD[next_number]) {
25        path.push(letter);
26        dfs(digits, path, res);
27        path.pop();
28    }
29}
30

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