Given a year .Your task is to find the number of every day in a year ie.number of Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday in that given year.
Examples:
Input: 2019
Output: Monday-52
Tuesday-53
Wednesday-52
Thursday-52
Friday-52
Saturday-52
Sunday-52
Input: 2024
Output: Monday-53
Tuesday-53
Wednesday-52
Thursday-52
Friday-52
Saturday-52
Sunday-52
Observations: We have to make some key observations. The first one will be that there are at least 52 weeks in a year, so every day will occur at least 52 times in a year. As 52*7 is 364 so the day occurring on the 1st January of any year will occur 53 times and if the year is a leap year then the day on the 2nd January will also occur 53 times.
Approach: Create a list with size 7 and having an initial value of 52 as the minimum number of occurrences will be 52. Find the index of the first day. Calculate the number of days whose occurrence will be 53.
Below is the implementation.
C++
#include <iostream>#include <ctime>#include <cstring>#include <cmath>using namespace std;void day_occur_time(int year) { // stores days in a week string days[] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" }; // Initialize all counts as 52 int L[7]; for (int i = 0; i < 7; i++) { L[i] = 52; } // Find the index of the first day // of the year int pos = -1; time_t t = time(NULL); tm* timePtr = localtime(&t); timePtr->tm_year = year - 1900; timePtr->tm_mon = 0; timePtr->tm_mday = 1; mktime(timePtr); char buffer[80]; strftime(buffer, 80, "%A", timePtr); string day(buffer); for (int i = 0; i < 7; i++) { if (day == days[i]) { pos = i; } } // mark the occurrence to be 53 of 1st day // and 2nd day if the year is leap year if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) { L[pos] += 1; L[(pos+1)%7] += 1; } else { L[pos] += 1; } // Print the days for (int i = 0; i < 7; i++) { cout << days[i] << " " << L[i] << endl; }}int main() { int year = 2019; day_occur_time(year); return 0;} |
Java
import java.time.LocalDate;import java.time.format.DateTimeFormatter;public class DayOccurTime { public static void dayOccurTime(int year) { // stores days in a week String[] days = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"}; // Initialize all counts as 52 int[] L = new int[7]; for (int i = 0; i < 7; i++) { L[i] = 52; } // Find the index of the first day of the year LocalDate date = LocalDate.of(year, 1, 1); String day = date.format(DateTimeFormatter.ofPattern("EEEE")); int pos = -1; for (int i = 0; i < 7; i++) { if (day.equals(days[i])) { pos = i; } } // mark the occurrence to be 53 of 1st day and 2nd day if the year is leap year if (date.isLeapYear()) { L[pos] += 1; L[(pos + 1) % 7] += 1; } else { L[pos] += 1; } // Print the days for (int i = 0; i < 7; i++) { System.out.println(days[i] + " " + L[i]); } } public static void main(String[] args) { int year = 2019; dayOccurTime(year); }}// This code is contributed by divyansh2212 |
Python3
# python program Find number of# times every day occurs in a Year import datetime import calendar def day_occur_time(year): # stores days in a week days = [ "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" ] # Initialize all counts as 52 L = [52 for i in range(7)] # Find the index of the first day # of the year pos = -1 day = datetime.datetime(year, month = 1, day = 1).strftime("%A") for i in range(7): if day == days[i]: pos = i # mark the occurrence to be 53 of 1st day # and 2nd day if the year is leap year if calendar.isleap(year): L[pos] += 1 L[(pos+1)%7] += 1 else: L[pos] += 1 # Print the days for i in range(7): print(days[i], L[i]) # Driver Code year = 2019day_occur_time(year) |
C#
using System;using System.Globalization;public class DayOccurTime{ public static void dayOccurTime(int year) { // stores days in a week string[] days = { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" }; // Initialize all counts as 52 int[] L = new int[7]; for (int i = 0; i < 7; i++) { L[i] = 52; } // Find the index of the first day of the year DateTime date = new DateTime(year, 1, 1); string day = date.ToString("dddd", CultureInfo.InvariantCulture); int pos = Array.IndexOf(days, day); // mark the occurrence to be 53 of 1st day and 2nd day if the year is leap year if (DateTime.IsLeapYear(year)) { L[pos] += 1; L[(pos + 1) % 7] += 1; } else { L[pos] += 1; } // Print the days for (int i = 0; i < 7; i++) { Console.WriteLine(days[i] + " " + L[i]); } } public static void Main(string[] args) { int year = 2019; dayOccurTime(year); }} |
Javascript
function day_occur_time(year) { // stores days in a week let days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" ]; // Initialize all counts as 52 let L = new Array(7).fill(52); // Find the index of the first day // of the year let pos = -1; let date = new Date(year, 0, 1); let day = date.toLocaleDateString('en-US', {weekday: 'long'}); for (let i = 0; i < 7; i++) { if (day == days[i]) { pos = i; } } // mark the occurrence to be 53 of 1st day // and 2nd day if the year is leap year if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) { L[pos] += 1; L[(pos+1)%7] += 1; } else { L[pos] += 1; } // Print the days for (let i = 0; i < 7; i++) { console.log(days[i] + " " + L[i]); }}let year = 2019;day_occur_time(year); |
Output
Monday 52 Tuesday 53 Wednesday 52 Thursday 52 Friday 52 Saturday 52 Sunday 52
Time complexity: O(1)
Auxiliary space: O(1)
