Problem

Given a date, return the corresponding day of the week for that date.

The input is given as three integers representing the day, month and year respectively.

Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}.

Examples

Example 1

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Input: day = 31, month = 8, year = 2019
Output: "Saturday"

Example 2

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Input: day = 18, month = 7, year = 1999
Output: "Sunday"

Example 3

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Input: day = 15, month = 8, year = 1993
Output: "Sunday"

Constraints

  • The given dates are valid dates between the years 1971 and 2100.

Solution

Method 1 – Zeller’s Congruence 1

Intuition

We can use Zeller’s Congruence, a well-known formula to compute the day of the week for any given date. It works by transforming the date into a formula that outputs the weekday as an integer, which we can map to the required weekday names.

Approach

  1. If the month is January or February, treat it as month 13 or 14 of the previous year.
  2. Apply Zeller’s Congruence formula to compute the weekday index.
  3. Map the index to the correct weekday name as per the problem’s requirements.
  4. Return the weekday name.

Code

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class Solution {
public:
    string dayOfTheWeek(int d, int m, int y) {
        vector<string> wd = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
        if (m < 3) {
            m += 12;
            y--;
        }
        int k = y % 100;
        int j = y / 100;
        int h = (d + 13 * (m + 1) / 5 + k + k / 4 + j / 4 + 5 * j) % 7;
        // Zeller's: 0=Saturday, 1=Sunday, ...
        return wd[(h + 6) % 7];
    }
};
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func dayOfTheWeek(d, m, y int) string {
    wd := []string{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}
    if m < 3 {
        m += 12
        y--
    }
    k := y % 100
    j := y / 100
    h := (d + 13*(m+1)/5 + k + k/4 + j/4 + 5*j) % 7
    return wd[(h+6)%7]
}
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class Solution {
    public String dayOfTheWeek(int d, int m, int y) {
        String[] wd = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
        if (m < 3) {
            m += 12;
            y--;
        }
        int k = y % 100;
        int j = y / 100;
        int h = (d + 13 * (m + 1) / 5 + k + k / 4 + j / 4 + 5 * j) % 7;
        return wd[(h + 6) % 7];
    }
}
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class Solution {
    fun dayOfTheWeek(d: Int, m: Int, y: Int): String {
        val wd = listOf("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday")
        var mm = m
        var yy = y
        if (mm < 3) {
            mm += 12
            yy--
        }
        val k = yy % 100
        val j = yy / 100
        val h = (d + 13 * (mm + 1) / 5 + k + k / 4 + j / 4 + 5 * j) % 7
        return wd[(h + 6) % 7]
    }
}
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class Solution:
    def dayOfTheWeek(self, d: int, m: int, y: int) -> str:
        wd = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
        if m < 3:
            m += 12
            y -= 1
        k = y % 100
        j = y // 100
        h = (d + 13 * (m + 1) // 5 + k + k // 4 + j // 4 + 5 * j) % 7
        return wd[(h + 6) % 7]
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impl Solution {
    pub fn day_of_the_week(d: i32, m: i32, y: i32) -> String {
        let wd = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"];
        let (mut m, mut y) = (m, y);
        if m < 3 {
            m += 12;
            y -= 1;
        }
        let k = y % 100;
        let j = y / 100;
        let h = (d + (13 * (m + 1)) / 5 + k + k / 4 + j / 4 + 5 * j) % 7;
        wd[((h + 6) % 7) as usize].to_string()
    }
}
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class Solution {
    dayOfTheWeek(d: number, m: number, y: number): string {
        const wd = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"];
        if (m < 3) {
            m += 12;
            y--;
        }
        const k = y % 100;
        const j = Math.floor(y / 100);
        const h = (d + Math.floor(13 * (m + 1) / 5) + k + Math.floor(k / 4) + Math.floor(j / 4) + 5 * j) % 7;
        return wd[(h + 6) % 7];
    }
}

Complexity

  • ⏰ Time complexity: O(1), as all operations are arithmetic and do not depend on input size.
  • 🧺 Space complexity: O(1), as only a few variables are used.