1118. Number of Days in a Month
Problem Description
The problem requires us to create a function that, given a year (year) and a month (month), returns the number of days in that given month. The challenge involves accounting for the different number of days in each month and determining whether the given year is a leap year since February has 29 days instead of 28 in leap years.
Intuition
The intuition behind the solution is to use a list to map each month to its respective number of days. We know that:
- January, March, May, July, August, October, and December all have 31 days.
- April, June, September, and November have 30 days.
- February has 28 days in a normal year and 29 days in a leap year.
To handle the special case of February in a leap year, we first need to determine whether the given year is a leap year. The determination can be made using the following rules:
- A year is a leap year if it is divisible by 4, except for end-of-century years, which must be divisible by 400.
- This means that if a year is divisible by 100 and not divisible by 400, it is NOT a leap year.
From these rules, we construct a logical condition that ensures the year is a leap year if it is either divisible by 400 or divisible by 4 but not by 100.
After this, we construct a list of days where February has 29 days if it's a leap year and 28 days otherwise. Then, the function simply returns the number of days corresponding to the given month by looking up the value in the pre-constructed list based on the month index.
Learn more about Math patterns.
In a binary min heap, the maximum element can be found in:
Solution Approach
The solution follows a simple and direct approach using a combination of condition checking and list indexing, which are fundamental constructs in programming - especially useful for this type of calendar-related computations.
We start by determining if the provided year is a leap year. The code does this with a single line of Boolean logic:
1leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
This line employs the modulus operator % to check for divisibility. The condition checks if the year is divisible by 4 but not by 100, unless it is also divisible by 400, in which case the year is indeed a leap year.
With the leap year status determined, we proceed to construct a list that maps each month (from index 1 to 12) to its corresponding number of days:
1days = [0, 31, 29 if leap else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
Notice that February (the second element of this list) has two possible values: 29 if leap is True, otherwise 28. The list starts with a placeholder 0 at index 0 since there is no month 0, and this alignment allows us to directly use the month value as an index to the list.
Finally, the function returns the number of days corresponding to the input month by accessing the days' list using the month as an index:
1return days[month]
The use of list indexing here provides an efficient and clean solution, avoiding multiple conditional statements or explicit date handling libraries, which are unnecessary for the problem at hand.
Which of the tree traversal order can be used to obtain elements in a binary search tree in sorted order?
Example Walkthrough
Let's walk through a small example to illustrate the solution approach. We'll create a function named get_days_in_month that implements the described approach. Suppose we are given the year 2020 which is a leap year and the month February, and we want to find out how many days February has in this year.
Given in 2020:
1year = 2020 2month = 2 # Since February is the second month
We first determine if 2020 is a leap year. The code:
1leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
evaluates to:
1leap = (2020 % 4 == 0 and 2020 % 100 != 0) or (2020 % 400 == 0)
As 2020 is divisible by 4 and not divisible by 100, leap becomes True. Since 2020 is a leap year, February will have 29 days.
Next, we create a list that maps each month to its number of days, accounting for leap years:
1days = [0, 31, 29 if leap else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
Which now translates to:
1days = [0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
because leap is True, index 2 (which corresponds to February) is assigned the value 29.
Finally, to find out the number of days in February 2020, the function performs a simple list indexing operation:
1return days[month]
Which gives us:
1return days[2]
The function returns 29, which is the correct number of days in February during a leap year.
So by following these steps, our function get_days_in_month would correctly determine that there are 29 days in February 2020. A straightforward sequence of logical checks and array indexing provides us with a concise and effective solution.
Solution Implementation
1class Solution:
2 def numberOfDays(self, year: int, month: int) -> int:
3 # Determine if the given year is a leap year. A year is a leap year if it is
4 # divisible by 4, but not by 100, unless it is also divisible by 400.
5 is_leap_year = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
6
7 # A list where the index corresponds to the month (1-12),
8 # and the value is the number of days in that month. February has
9 # 29 days if it's a leap year, or 28 days if it's not.
10 days_in_month = [
11 0, # Index 0 - not used, for easier matching of month to index
12 31, # January
13 29 if is_leap_year else 28, # February
14 31, # March
15 30, # April
16 31, # May
17 30, # June
18 31, # July
19 31, # August
20 30, # September
21 31, # October
22 30, # November
23 31 # December
24 ]
25
26 # Return the number of days in the specified month.
27 return days_in_month[month]
281class Solution {
2
3 /**
4 * Calculates the number of days in a given month for a specific year
5 *
6 * @param year The year as an integer value
7 * @param month The month as an integer value
8 * @return The number of days in the given month of the year
9 */
10 public int numberOfDays(int year, int month) {
11 // Determine if the year is a leap year
12 boolean isLeapYear = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
13
14 // Array holding the number of days in each month; for February, use leap year value if applicable
15 int[] daysPerMonth = new int[] {
16 0, // Placeholder for indexing purposes; there is no month 0
17 31, // January
18 isLeapYear ? 29 : 28, // February
19 31, // March
20 30, // April
21 31, // May
22 30, // June
23 31, // July
24 31, // August
25 30, // September
26 31, // October
27 30, // November
28 31 // December
29 };
30
31 // Return the number of days in the specified month
32 return daysPerMonth[month];
33 }
34}
351#include <vector> // Include the vector header for using the vector container
2
3class Solution {
4public:
5 // Function to determine the number of days in a given month of a given year
6 int numberOfDays(int year, int month) {
7 // Check if the year is a leap year; a leap year is divisible by 4, not divisible by 100 unless also divisible by 400
8 bool isLeapYear = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
9
10 // Initialize a vector with the number of days in each month; February has 29 days if it is a leap year, otherwise 28
11 std::vector<int> daysInMonth = {0, 31, isLeapYear ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
12
13 // Return the number of days in the specified month
14 return daysInMonth[month];
15 }
16};
171// This function calculates the number of days in a given month for a specified year.
2// It accounts for leap years when determining the number of days in February.
3function numberOfDays(year: number, month: number): number {
4 // Check if the year is a leap year. A year is a leap year if it is divisible by 4
5 // but not by 100, or if it is divisible by 400.
6 const isLeapYear: boolean = (year % 4 === 0 && year % 100 !== 0) || year % 400 === 0;
7
8 // Create an array representing the number of days in each month.
9 // For February (index 2), use 29 days if it's a leap year, otherwise use 28.
10 const daysPerMonth: number[] = [0, 31, isLeapYear ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
11
12 // Return the number of days of the specified month.
13 return daysPerMonth[month];
14}
15A person thinks of a number between 1 and 1000. You may ask any number questions to them, provided that the question can be answered with either "yes" or "no".
What is the minimum number of questions you needed to ask so that you are guaranteed to know the number that the person is thinking?
Time and Space Complexity
Time Complexity
The time complexity of the given code is O(1) because it performs a constant number of operations no matter the value of the input year and month. Checking if a year is a leap year and accessing an element from a pre-defined list both take constant time.
Space Complexity
The space complexity of the code is also O(1) as it uses a fixed amount of additional memory. The list days is of a constant size (13 elements), and the space required does not grow with the size of the input year or month.
Learn more about how to find time and space complexity quickly using problem constraints.
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