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)

// 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<string> names = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
    vector<int> 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];
  }
};

The post 花花酱 LeetCode 1185. Day of the Week appeared first on Huahua's Tech Road.

Example 1:

Input: date = "2019-01-09"
Output: 9
Explanation: Given date is the 9th day of the year in 2019.

Example 2:

Input: date = "2019-02-10"
Output: 41

Example 3:

Input: date = "2003-03-01"
Output: 60

Example 4:

Input: date = "2004-03-01"
Output: 61

Constraints:

• date.length == 10
• date[4] == date[7] == '-', and all other date[i]‘s are digits
• date represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.

Solution:

Key: checking whether that year is a leap year or not.
is_leap = (year % 4 == 0 and year % 100 !=0) or year % 400 == 0

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

# Author: Huahua
class Solution:
  def dayOfYear(self, date: str) -> int:
    y, m, d = int(date.split("-")[0]), int(date.split("-")[1]), int(date.split("-")[2])
    leap = (y % 4 == 0 and y % 100 != 0) or y % 400 == 0
    days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    s = sum(days[0:m-1]) + d
    if leap and m > 2: s += 1
    return s

The post 花花酱 LeetCode 1154. Day of the Year appeared first on Huahua's Tech Road.