date Archives - Huahua's Tech Road

花花酱 LeetCode 2409. Count Days Spent Together

zxi — Sat, 17 Sep 2022 19:55:26 +0000

Alice and Bob are traveling to Rome for separate business meetings.

You are given 4 strings arriveAlice, leaveAlice, arriveBob, and leaveBob. Alice will be in the city from the dates arriveAlice to leaveAlice (inclusive), while Bob will be in the city from the dates arriveBob to leaveBob (inclusive). Each will be a 5-character string in the format "MM-DD", corresponding to the month and day of the date.

Return the total number of days that Alice and Bob are in Rome together.

You can assume that all dates occur in the same calendar year, which is not a leap year. Note that the number of days per month can be represented as: [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31].

Example 1:

Input: arriveAlice = "08-15", leaveAlice = "08-18", arriveBob = "08-16", leaveBob = "08-19"
Output: 3
Explanation: Alice will be in Rome from August 15 to August 18. Bob will be in Rome from August 16 to August 19. They are both in Rome together on August 16th, 17th, and 18th, so the answer is 3.

Example 2:

Input: arriveAlice = "10-01", leaveAlice = "10-31", arriveBob = "11-01", leaveBob = "12-31"
Output: 0
Explanation: There is no day when Alice and Bob are in Rome together, so we return 0.

Constraints:

All dates are provided in the format "MM-DD".
Alice and Bob’s arrival dates are earlier than or equal to their leaving dates.
The given dates are valid dates of a non-leap year.

Solution: Math

Convert date to days of the year.

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  int countDaysTogether(string arriveAlice, string leaveAlice, string arriveBob, string leaveBob) {
    array days{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    auto getDay = [&](string date) {
      int mm = (date[0] - '0') * 10 + (date[1] - '0');
      int dd = (date[3] - '0') * 10 + (date[4] - '0');
      return accumulate(begin(days), begin(days) + mm - 1, dd);
    };
    
    int s = max(getDay(arriveAlice), getDay(arriveBob));
    int e = min(getDay(leaveAlice), getDay(leaveBob));
    return e >= s ? e - s + 1 : 0;
  }
};

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花花酱 LeetCode 1751. Maximum Number of Events That Can Be Attended II

zxi — Sat, 06 Feb 2021 16:37:38 +0000

You are given an array of events where events[i] = [startDay_i, endDay_i, value_i]. The i^th event starts at startDay_iand ends at endDay_i, and if you attend this event, you will receive a value of value_i. You are also given an integer k which represents the maximum number of events you can attend.

You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day.

Return the maximum sum of values that you can receive by attending events.

Example 1:

Input: events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
Output: 7
Explanation: Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.

Example 2:

Input: events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
Output: 10
Explanation: Choose event 2 for a total value of 10.
Notice that you cannot attend any other event as they overlap, and that you do not have to attend k events.

Example 3:

Input: events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
Output: 9
Explanation: Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.

Constraints:

1 <= k <= events.length
1 <= k * events.length <= 10⁶
1 <= startDay_i <= endDay_i <= 10⁹
1 <= value_i <= 10⁶

Solution: DP + Binary Search

Sort events by ending time.
dp[i][j] := max value we can get by attending at most j events among events[0~i].
dp[i][j] = max(dp[i – 1][j], dp[p][j – 1] + value[i])
p is the first event that does not overlap with the current one.

Time complexity: O(nlogn + nk)
Space complexity: O(nk)

C++

// Author: Huahua
class Solution {
public:
  int maxValue(vector>& events, int k) {
    const int n = events.size(); 
    vector> dp(n + 1, vector(k + 1));    
    vector e(n);
    auto comp = [](const vector& a, const vector& b) {
      return a[1] < b[1];
    };
    
    sort(begin(events), end(events), comp);
    
    for (int i = 1; i <= n; ++i) {
      const int p = lower_bound(begin(events), 
                                begin(events) + i, 
                                vector{0, events[i - 1][0], 0},
                                comp) - begin(events);
      for (int j = 1; j <= k; ++j)
        dp[i][j] = max(dp[i - 1][j], 
                       dp[p][j - 1] + events[i - 1][2]);
    }    
    return dp[n][k];
  }
};

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花花酱 LeetCode 1507. Reformat Date

zxi — Sun, 12 Jul 2020 03:58:04 +0000

Given a date string in the form Day Month Year, where:

Day is in the set {"1st", "2nd", "3rd", "4th", ..., "30th", "31st"}.
Month is in the set {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"}.
Year is in the range [1900, 2100].

Convert the date string to the format YYYY-MM-DD, where:

YYYY denotes the 4 digit year.
MM denotes the 2 digit month.
DD denotes the 2 digit day.

Example 1:

Input: date = "20th Oct 2052"
Output: "2052-10-20"

Example 2:

Input: date = "6th Jun 1933"
Output: "1933-06-06"

Example 3:

Input: date = "26th May 1960"
Output: "1960-05-26"

Constraints:

The given dates are guaranteed to be valid, so no error handling is necessary.

Solution: String + HashTable

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  string reformatDate(string date) {
    stringstream ss(date);
    string day, month, year;
    ss >> day >> month >> year;
    unordered_map m{{"Jan", "01"},
                                    {"Feb", "02"},
                                    {"Mar", "03"},
                                    {"Apr", "04"},
                                    {"May", "05"},
                                    {"Jun", "06"},
                                    {"Jul", "07"},
                                    {"Aug", "08"},
                                    {"Sep", "09"},
                                    {"Oct", "10"},
                                    {"Nov", "11"},
                                    {"Dec", "12"}};
    day = day.substr(0, day.length() - 2);
    if (day.length() == 1) day = "0" + day;
    return year + "-" + m[month] + "-" + day;
  }
};

Java

// Author: Huahua
class Solution {
  public String reformatDate(String date) {
    Map m = new HashMap();
    m.put("Jan", "01");
    m.put("Feb", "02");
    m.put("Mar", "03");
    m.put("Apr", "04");
    m.put("May", "05");
    m.put("Jun", "06");
    m.put("Jul", "07");
    m.put("Aug", "08");
    m.put("Sep", "09");
    m.put("Oct", "10");
    m.put("Nov", "11");
    m.put("Dec", "12");
    String[] items = date.split(" ");    
    String day = items[0].substring(0, items[0].length() - 2);
    if (day.length() == 1) day = "0" + day;
    return items[2] + "-" + m.get(items[1]) + "-" + day;
  }
}

Python

# Author: Huahua
class Solution:
  def reformatDate(self, date: str) -> str:
    m = {"Jan": "01", "Feb": "02", "Mar": "03", 
         "Apr": "04", "May": "05", "Jun": "06", 
         "Jul": "07", "Aug": "08", "Sep": "09", 
         "Oct": "10", "Nov": "11", "Dec": "12"}
    items = date.split(" ")
    day = items[0][:-2]
    if len(day) == 1: day = "0" + day
    return items[2] + "-" + m[items[1]] + "-" + day

The post 花花酱 LeetCode 1507. Reformat Date appeared first on Huahua's Tech Road.

花花酱 LeetCode 1360. Number of Days Between Two Dates

zxi — Sun, 23 Feb 2020 09:20:02 +0000

Write a program to count the number of days between two dates.

The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples.

Example 1:

Input: date1 = "2019-06-29", date2 = "2019-06-30"
Output: 1

Example 2:

Input: date1 = "2020-01-15", date2 = "2019-12-31"
Output: 15

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Solution: Convert to days since epoch

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  int daysBetweenDates(string date1, string date2) {
    array m{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    auto isLeap = [](int year) {
      return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
    };
    auto daysFromEpoch = [&](const string& date) {
      int year = stoi(date.substr(0, 4));
      int month = stoi(date.substr(5, 2));
      int days = stoi(date.substr(8, 2));      
      for (int i = 1970; i < year; ++i)
        days += 365 + isLeap(i);
      for (int i = 1; i < month; ++i)
        days += m[i - 1];
      days += month > 2 && isLeap(year);
      return days;
    };
    return abs(daysFromEpoch(date1) - daysFromEpoch(date2));
  }
};

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花花酱 LeetCode 1185. Day of the Week

zxi — Sun, 08 Sep 2019 04:18:17 +0000

Given a date, return the corresponding day of the week for that date.

The input is given as three integers representing the day, month and year respectively.

Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}.

Example 1:

Input: day = 31, month = 8, year = 2019
Output: "Saturday"

Example 2:

Input: day = 18, month = 7, year = 1999
Output: "Sunday"

Example 3:

Input: day = 15, month = 8, year = 1993
Output: "Sunday"

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Solution: LeapYear

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua
inline bool leapYear(int y) {
  return (y % 4 == 0 && y % 100 != 0) || (y % 400 == 0);
}
class Solution {
public:
  string dayOfTheWeek(int day, int month, int year) {    
    vector names = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
    vector days = {31, 28 + leapYear(year), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int sum = 0;
    for (int i = 1970; i < year; ++i)
      sum += 365 + leapYear(i);    
    for (int i = 1; i < month; ++i)
      sum += days[i - 1];    
    sum += day;
    return names[(sum + 3) % 7];
  }
};

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