Day Name
Given the weekday of the first day of the month, determine the day of the week of the given date in that month.
Input
The first line is a string D.
The second line is an integer N.
Output
The output should be a string.
Explanation
In the given example, D = Monday. As the 1st of the day of the month is a Monday, it means the 7th and 14th of the month will also be Mondays (A week has 7 days). So the 16th day (N = 16) of the month will be a Tuesday.
So, the output should be Tuesday.
Sample Input 1:
Monday
16
Sample Output 1:
Tuesday
Sample Input 2:
Tuesday
17
Sample Output 2:
Thursday
Approach:
def determine_day_by_number(start_day: str, month_day_number: int):
number_to_day = dict(
[[1, 'Monday'], [2, 'Tuesday'], [3, 'Wednesday'], [4, 'Thursday'], [5, 'Friday'], [6, 'Saturday'],
[7, 'Sunday']])
day_to_number = dict(zip(number_to_day.values(), number_to_day.keys()))
if start_day not in number_to_day.values():
print('Wrong day name')
exit()
start_day_number = day_to_number[start_day]
shift = get_day_shift(start_day_number, month_day_number)
number_of_required_day = start_day_number + shift
return number_to_day[number_of_required_day]
def get_day_shift(start_day_number: int, month_day_number: int):
shift = month_day_number
if start_day_number + shift > 7:
number_of_full_weeks = shift // 7
shift -= 7 * number_of_full_weeks
elif start_day_number == 1:
return start_day_number
return shift if not start_day_number + shift > 7 else shift - 7
start_day = input()
month_day_number = int(input())
day = determine_day_by_number(start_day, month_day_number)
print(day)
INPUT:
Tuesday
1
Output should be:
Tuesday
My Output:
Wednesday
def determine_day_by_number(start_day: str, month_day_number: int):
number_to_day = dict(
[[1, 'Monday'], [2, 'Tuesday'], [3, 'Wednesday'], [4, 'Thursday'], [5, 'Friday'], [6, 'Saturday'],
[7, 'Sunday']])
day_to_number = dict(zip(number_to_day.values(), number_to_day.keys()))
if start_day not in number_to_day.values():
print('Wrong day name')
exit()
start_day_number = day_to_number[start_day]
shift = get_day_shift(start_day_number, month_day_number)
number_of_required_day = start_day_number + shift
return number_to_day[number_of_required_day]
def get_day_shift(start_day_number: int, month_day_number: int):
shift = month_day_number
if start_day_number + shift > 7:
number_of_full_weeks = shift // 7
shift -= 7 * number_of_full_weeks + 1
return shift if not start_day_number + shift > 7 else shift - 7
def main():
start_day = input()
month_day_number = int(input())
day = determine_day_by_number(start_day, month_day_number)
print(day)
if __name__ == '__main__':
main()
dayseq = ["Sunday", "Monday", "Tuesday",
"Wednesday", "Thursday", "Friday", "Saturday"]
inputday = input("Enter a day: ") # monday
inputdate = int(input("Enter a date for the above day: ")) # 4
inputdayofdate = int(input("Enter a date to find day: ")) # 7 Thursday
step = 1
def loopers(currentday, step):
if step == -1:
if currentday == "Sunday":
return "Saturday"
return dayseq[dayseq.index(currentday)-1]
elif step == 1:
if currentday == "Saturday":
return "Sunday"
return dayseq[dayseq.index(currentday)+1]
if inputdayofdate < inputdate:
step = -1
for i in range(inputdate, inputdayofdate, step):
inputday = loopers(inputday, step)
print(inputday)
The above code will take 3 inputs (day date anddatetofind) and it will calculate day from the set day and date
OR
dayseq = ["Sunday", "Monday", "Tuesday",
"Wednesday", "Thursday", "Friday", "Saturday"]
inputday = input("Enter a day: ") # monday
inputdayofdate = int(input("Enter a date to find day: ")) # 7 Thursday
def loopers(currentday):
if currentday == "Saturday":
return "Sunday"
return dayseq[dayseq.index(currentday)+1]
for i in range(inputdayofdate -1):
inputday = loopers(inputday)
print(inputday)
The above code with consider given day as 1st of a month and continue
I was solving Project Euler #19:
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
And here is the code :
months = { "January": 31,
"February" : 28,
"March" : 31,
"April" : 30,
"May" : 31,
"June" : 30,
"July" : 31,
"August" : 31,
"September" : 30,
"October" : 31,
"November" : 30,
"December" : 31}
def countingSundays():
day = 1
sunday_count = 0
for year in xrange(1901,2001):
for m in months:
day += months[m]
if year % 4 == 0 and m == "February":
day += 1
if day % 7 == 0:
sunday_count += 1
print "Sundays:", sunday_count
The output of the program is 172 which is incorrect.
I searched the answer to be 171.
So I wanted to know why am I getting the extra 1 Sunday ?
You're iterating over the months dict, expecting it to iterate in the order of the months, but dicts aren't ordered, so you can get the months in the wrong order.
Since you don't actually need the month names, you can just make months a list of the month lengths instead.
You should use the datetime library, which will handled all the leap year information automatically:
from datetime import date
from collections import Counter
counter = Counter()
for year in xrange(1901, 2001):
for month in xrange(1, 13):
day = date(year, month, 1)
counter[day.weekday()] += 1
print counter[6]
import time
from math import floor
"""
Gaussian algorithm to determine day of week
"""
def day_of_week(year, month, day):
"""
w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7
Y = year - 1 for January or February
Y = year for other months
d = day (1 to 31)
m = shifted month (March = 1, February = 12)
y = last two digits of Y
c = first two digits of Y
w = day of week (Sunday = 0, Saturday = 6)
"""
d = day
m = (month - 3) % 12 + 1
if m > 10: Y = year - 1
else: Y = year
y = Y % 100
c = (Y - (Y % 100)) / 100
w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7
return int(w)
"""
Compute the number of months starting on a given day of the week in a century
"""
def months_start_range(day,year_start,year_end):
total = 0
for year in range(year_start, year_end + 1):
for month in range(1,13):
if day_of_week(year, month, 1) == day: total += 1
return total
start = time.time()
total = months_start_range(0,1901,2000)
elapsed = time.time() - start
print("%s found in %s seconds") % (total,elapsed)
This might you solve the problem.
It took around 0.068 seconds to solve it.
Here is a different approach to tackle this question
public static void main(String[] args) {
int k = 0;
// String months[] = { "January", "February", "March", "April", "May", "June",
// "July", "August", "September",
// "October", "November", "December" };
String Days[] = { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" };
int MonthsDdays[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int counter = 0;
for (int t = 1900; t <= 2000; t++) {
MonthsDdays[1]=28;
if (t % 4 == 0) {
if (t % 100 == 0)
{
if (t % 400 == 0)
MonthsDdays[1] = 29;
} else if (t % 100 != 0)
MonthsDdays[1] = 29;
}
int p = 0;
while (p < 12) {
for (int j = 0; j < MonthsDdays[p]; k++, j++) {
if (k == 7)
k = 0;
if (Days[k].equalsIgnoreCase("Sunday") && j == 0 && t > 1900) {
counter++;
}
}
p++;
}
}
System.out.println(counter);
}
I tried the Mathematical approach although we could use the calendar functions.
I first calculated the math of the months to determine the relationships between the first dates of the months using the other months. Also, for simplicity in calculating leap years, I calculated the year from March to Feb. If you want to calculate for the Jan and Feb of 1901, you can write a separate condition, and do the same to remove Jan and Feb of 2001. However, in this case, they do not really matter as they are not Sundays, so you could remove the last if condition for this specific case.
# Zero is Sunday and the rest of the days are according to mod7
# li stores the first days of the months in the year in every iteration
# The year in initial li is 1900 but the contents are Mar-1900 to Feb-1901
# At the end, we can check if Jan or Feb of 2001 contain a Sunday and remove if it does
li, cnt = [4,0,2,5,0,3,6,1,4,6,2,5], 0
# Could also initialize li from by the same method as below, but I had already calculated those
# cnt adds number of zeros in every iteration (the number of Sundays in every year) to its value
# As we don't count for the year 1900 cnt=0, else initialize cnt=li.count(0)
for year in range(1901,2001):
if year%4==0:
li[0]=li[8]=(li[11]+1)%7 #Set March and November to +1 value than last Feb
else:
li[0]=li[8]=li[11] #Set March and November to same value as last Feb
# The following values of other months will depend solely on their own March value
# You can check the Math if you want to
li[3]=li[11]=(li[0]+1)%7;li[6]=li[9]=(li[0]+2)%7;li[1]=li[4]=(li[0]+3)%7;li[2]=li[10]=(li[0]-2)%7;li[5]=(li[0]-1)%7;li[7]=(li[0]-3)%7
cnt = cnt + li.count(0)
# This is to remove the extra two months of the year 2001 if they bother the answer
if li[10] == 0 or li[11] == 0:
cnt = cnt-1
print(cnt)
This was my first answer on StackOverflow, I hope I wrote well. ;-)
The mistakes you have:
The way you calculate leap years
Dictionary does not keep the order necessarily
You assume January 1st is Sunday
The correct program would be:
from collections import OrderedDict
months = OrderedDict( [("January",31),("February", 28),("March",31),
("April", 30), ("May", 31), ("June", 30),
("July", 31), ("August", 31), ("September", 30),
("October", 31), ("November", 30), ("December", 31)] )
days = ['Tuesday','Wednesday', 'Thursday','Friday','Saturday', 'Sunday', 'Monday']
day = 0
sunday_count = 0
def isLeap(year): #https://en.wikipedia.org/wiki/Leap_year#Algorithm
leap = True
if year % 4 != 0:
leap = False
elif year % 100 != 0:
leap = True
elif year % 400 != 0:
leap = False
return leap
for year in xrange(1901,2001):
leap = isLeap(year)
for m in months:
dayName = days[day%7]
if dayName == "Sunday":
sunday_count += 1
#print year, m, dayName
day += months[m]
if leap == True and m == "February":
day += 1
print sunday_count
# print 171
Also, some days:
1901 January Tuesday
1901 February Friday
1901 March Friday
1901 April Monday
1901 May Wednesday
1901 June Saturday
1901 July Monday
1901 August Thursday
1901 September Sunday
...
import pandas as pd
from datetime import date
start = date(1901, 1, 1)
end = date(2000, 12, 31)
d = pd.date_range(start, end, freq='MS').strftime('%A')
s = pd.Series(d)
print(s.value_counts())
So I approached this problem from not a date perspective but of a counting days.
Here's my solution:.
days_1st = list()
day_counter = 1
for year in range(1900, 2001):
for month in range(1,13):
#Skip for year 1900 as count starts from 1901, but this still
#adds the days hence keeping the cycle in sync!
if year != 1900:
days_1st.append(day_counter)
if month == 4 or month == 6 or month == 9 or month == 11:
day_counter+=30
elif month == 2 and ((year % 100 == 0 and year % 400 == 0) or (year % 100 != 0 and year % 4 == 0)):
day_counter+=29
elif month == 2:
day_counter+=28
else:
day_counter+=31
# mod 7 because since the day the counting started (1 Jan 1900 -
# Monday) Every 7th day is a sunday!
days_sunday = list(filter(lambda x: x % 7 == 0, days_1st))
print(len(days_sunday))
A = [31,28,31,30,31,30,31,31,30,31,30,31]
sunday =0
gK = 1
for y in range(1901,2001):
if(y %4 ==0):
A[1] = 29
else:
A[1] = 28
for m in range(len(A)):
for d in range(1,A[m]+1):
if(gK ==6):
if(d==1):
sunday +=1
gK =0
else:
gK =gK+1
print(sunday)
==>Solution in python
euler19.py
normal_year = [31,28,31,30,31,30,31,31,30,31,30,31]
leap_year = [31,29,31,30,31,30,31,31,30,31,30,31]
years = [ normal_year ] * 100
for i in range(3, len(years), 4) :
years[i] = leap_year
current_day = (0+365) % 7
sundays = 0
for y in years :
for m in y :
if current_day % 7 == 6:
sundays += 1
current_day += m%7
print (sundays)
I think I got the answer. I am not sure though.. your logic was right. But needed a little improvement. We need to start off by counting the number of Tuesdays first as we clearly know that it was Monday on Jan 1, 1900.
months = { "January": 31,
"February" : 28,
"March" : 31,
"April" : 30,
"May" : 31,
"June" : 30,
"July" : 31,
"August" : 31,
"September" : 30,
"October" : 31,
"November" : 30,
"December" : 31}
for month in months:
print(months[month])
tuesday_count = 0
day = 0
extra_days = 0
for year in range(1901, 2001):
days_in_the_year = 0
for month in months:
day += months[month]
days_in_the_year += months[month]
if( year % 4 == 0 and month == 'February'):
if (year % 100 != 0):
extra_days += 1
days_in_the_year += 1
day += 1
elif(year % 100 ==0 and year % 400 ==0):
extra_days += 1
days_in_the_year += 1
day += 1
if( (day) % 7 == 0):
tuesday_count += 1
print('No. of days in the year',year,'are',days_in_the_year)
print('No. of Tuesdays counted so far is =', tuesday_count)
print('The number of extra_days because of the leap years are:',extra_days)
# print('extra_days % 7 =', '25 % 7 =', extra_days % 7)
print('So, there were', extra_days // 7, 'extra_no_of_weeks left that we haven\'t considered. After that, it\'s followed by --wed, thu, fri and sat (we don\'t need to consider that).\n So, the total number of Tuesdays are', tuesday_count+3 )
tuesday_count += 3
print('This means only 2 Sundays that have followed')
sunday_count = tuesday_count - 1
print('Since, 1901 Jan1 would be a Tuesday, we need to subract one from the total number of Sundays\n So, the total number of sundays are:', )
sunday_count=sunday_count-1
print(sunday_count)
months=[31,28,31,30,31,30,31,31,30,31,30,31]
leap=[31,29,31,30,31,30,31,31,30,31,30,31]
sundays=0
start=2
for y in range(25):
for nonleap in range (3):
for j in months:
start=(start+j)%7
if start == 0:
sundays+=1
for m in leap:
start=(start+m)%7
if start == 0:
sundays+=1
print sundays
Note that the problem defines the first day of 1900 as Monday and you define the first day of 1901 as Monday.
months = [31,28,31,30,31,30,31,31,30,31,30,31]
def countingSundays():
day = 1
sunday_count = 0
for year in range(1900,1901):
for m in months:
day += m
if (year % 4 == 0 and m == 28):
day += 1
for year in range(1901,2001):
for m in months:
day += m
if (year % 4 == 0 and m == 28):
day += 1
if day % 7 == 0:
sunday_count += 1
return sunday_count
print ("Sundays:", countingSundays())
you have initialized the day variable to 1 but the 1st Jan 1901 is a Tuesday. I made the same error ;-)