This question was already done by me earlier. So, just thought of already built logic and wrote the code. Initially, few test cases were not passed then I made some changes, and all test cases passed.

Resources - https://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week

 

There are multiple ways and complicated algorithms to solve this problem like Tomohiko Sakamoto’s Algorithm and Zeller Formula. You can look for them over the web, but we will go with a more intuitive solution.

If we know today is Sunday then we can easily tell that after 2 days it will be Tuesday or after 10 days it will be Wednesday.

Similarly, we know according to the Gregorian calendar (the calendar we follow) that it’s Monday on 1st Jan 1 AD. So, if we can count the number of days passed till current day then we can easily find the Day of the Week for the current day.

To make our calculations easier we can first skip leap days and calculate the number of days passed. Then we can add the leap days separately.

Also, we can precalculate and keep some constant values like:

• Array Storing Names of Days in the correct order of Modulo 7.

DaysName[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};

• Number of Days Passed till Month 0,1,2....11 ignoring leap years.

noOfDaysPassedTillMonth[] = {0,31,59,90,120,151,181,212,243,273,304,334}

So ignoring leap days, totalNoOfDaysPassed = 365*(year - 1) + noOfDaysPassedTillMonth[month-1] + days.

Now for leap days, we need to count the number of leap years passed till current year also we need to ignore the current year if leap day i.e. 29th February is not passed in the current year. 

Pseudocode:

if(month<=2)

    year--

Then, we can calculate the number of leap years by counting the number of multiples of 4 excluding the number of multiples of 100 and finally including the number of multiples of 400. 

Pseudocode:

leapDays = (year/4) - (year/100) + (year/400);

totalNoOfDaysPassed += leapDays;

return DaysName[totalNoOfDaysPassed % 7]

O(1).

As we are just using constant space to store some variables irrespective of input date.

O(1).

As we are just doing some contact calculations irrespective of input date.