Day of the Week

Intuition

The cleanest way to think about this problem is:

• count how many days have passed since January 1, 1971
• remember that January 1, 1971 was a Friday
• use modulo 7 to map the total back to the correct weekday

Once we reduce the question to "how many days passed before this date?", the rest becomes simple bookkeeping.

Approach

We keep two precomputed arrays:

• years[i] stores how many days have passed from 1971 to 1971 + i
• months[i] stores how many days have passed before month i + 1 in a normal year

Then we build the answer in four steps:

1. Add the days contributed by full years before year
2. Add the days contributed by full months before month
3. Add the current day
4. If the year is a leap year and the month is after February, add one extra day

At that point we know the total offset from the reference Friday, so the answer is:

days[(totalDays - 1) % 7]

class Solution {
    public String dayOfTheWeek(int day, int month, int year) {
        int[] years = {0,365,731,1096,1461,1826,2192,2557,2922,3287,3653,4018,4383,4748,5114,5479,5844,6209,6575,6940,7305,7670,8036,8401,8766,9131,9497,9862,10227,10592,10958,11323,11688,12053,12419,12784,13149,13514,13880,14245,14610,14975,15341,15706,16071,16436,16802,17167,17532,17897,18263,18628,18993,19358,19724,20089,20454,20819,21185,21550,21915,22280,22646,23011,23376,23741,24107,24472,24837,25202,25568,25933,26298,26663,27029,27394,27759,28124,28490,28855,29220,29585,29951,30316,30681,31046,31412,31777,32142,32507,32873,33238,33603,33968,34334,34699,35064,35429,35795,36160,36525,36890,37256,37621,37986,38351,38717,39082,39447,39812,40178,40543,40908,41273,41639,42004,42369,42734,43100,43465,43830,44195,44561,44926,45291,45656,46022,46387,46752,47117};
        int[] months = {0,31,59,90,120,151,181,212,243,273,304,334,365};
        String[] days = {"Friday", "Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday"};
        int totalDays = 0;

        totalDays += years[year - 1971];
        totalDays += months[month - 1];

        if (isLeap(year) && month > 2) {
            totalDays++;
        }

        totalDays += day;
        return days[(totalDays - 1) % 7];
    }

    private boolean isLeap(int year) {
        return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    }
}

Leap Year Rule

A year is a leap year if:

• it is divisible by 4 and not divisible by 100, or
• it is divisible by 400

That is why 2000 is a leap year, but 1900 is not.

Dry Run

Take:

day = 31
month = 8
year = 2019

Now compute:

• days from full years before 2019
• days from months before August
• add 31
• 2019 is not a leap year, so no extra day

After taking modulo 7, we land on Saturday.

Complexity

Time Complexity: O(1)

Space Complexity: O(1)

Why This Works Well

This solution avoids simulating calendars month by month or year by year.

Once the reference date is fixed, the whole problem becomes a direct index calculation. That makes it fast, easy to reason about, and very reliable during interviews.