Find day of the week for a given date

Write a function that calculates the day of the week for any particular date in the past or future. A typical application is to calculate the day of the week on which someone was born or some other special event occurred. 

Following is a simple function suggested by Sakamoto, Lachman, Keith and Craver to calculate day. The following function returns 0 for Sunday, 1 for Monday, etc. 

Understanding the Maths:

14/09/1998
dd=14
mm=09
yyyy=1998 //non-leap year

Step 1: Informations to be remembered.
 Magic Number Month array.
 For Year: {0,3,3,6,1,4,6,2,5,0,3,5}
 DAY array starting from 0-6: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
 Century Year Value: 1600-1699 = 6
      1700-1799 = 4
      1800-1899 = 2
      1900-1999 = 0
      2000-2099 = 6..

Step 2: Calculations as per the steps
 
 Last 2 digits of the year:  98
 Divide the above by 4:     24
 Take the date(dd):      14
 Take month value from array: 5 (for September month number 9)
 Take century year value:  0  ( 1998 is in the range 1900-1999 thus 0)
          -----
 Sum:        141
 
 Divide the Sum by 7 and get the remainder: 141 % 7 = 1
 
 Check the Day array starting from index 0: Day[1] = Monday

**If leap year it will be the remainder-1

C++

/* A program to find day of a given date */
#include <bits/stdc++.h>
using namespace std;

int dayofweek(int d, int m, int y)
{
    static int t[] = { 0, 3, 2, 5, 0, 3,
                       5, 1, 4, 6, 2, 4 };
    y -= m < 3;
    return ( y + y / 4 - y / 100 +
             y / 400 + t[m - 1] + d) % 7;
}

// Driver Code
int main()
{
    int day = dayofweek(30, 8, 2010);
    cout << day;

    return 0;
}

// This is code is contributed
// by rathbhupendra

C

/* A program to find day of a given date */
#include<stdio.h>

int dayofweek(int d, int m, int y)
{
    static int t[] = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 };
    y -= m < 3;
    return ( y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}

/* Driver function to test above function */
int main()
{
    int day = dayofweek(30, 8, 2010);
    printf ("%d", day);

    return 0;
}

Java

/*This code will return the string value and the exact date
 * both for leap and non leap years*/
import java.util.*;
class FindDay {
    static int dd;
    static int mm;
    static int yyyy;
    FindDay(int dd, int mm, int yyyy)
    {
        this.dd = dd;
        this.mm = mm;
        this.yyyy = yyyy;
    }
    static int checkLeap(int y)
    {
        if ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0))
            return 1;
        else
            return 0;
    }
    static void calculate()
    {

        // Checking Leap year. If true then 1 else 0.
        int flag_for_leap = checkLeap(yyyy);

        /*Declaring and initialising the given informations
         * and arrays*/
        String day[] = { "Sunday",    "Monday",   "Tuesday",
                         "Wednesday", "Thursday", "Friday",
                         "Saturday" };
        int m_no[] = { 0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5 };

        /*Generalised check to find any Year Value*/
        int j;
        if ((yyyy / 100) % 2 == 0) {
            if ((yyyy / 100) % 4 == 0)
                j = 6;
            else
                j = 2;
        }
        else {
            if (((yyyy / 100) - 1) % 4 == 0)
                j = 4;
            else
                j = 0;
        }

        /*THE FINAL FORMULA*/
        int total = (yyyy % 100) + ((yyyy % 100) / 4) + dd
                    + m_no[mm - 1] + j;
        if (flag_for_leap == 1) {
            if ((total % 7) > 0)
                System.out.println(day[(total % 7) - 1]);
            else
                System.out.println(day[6]);
        }
        else
            System.out.println(day[(total % 7)]);
    }
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        /*Take input in string format and then convert to
         * integer values*/
        String date = sc.next();
        String[] values = date.split("/");
        new FindDay(Integer.parseInt(values[0]),
                    Integer.parseInt(values[1]),
                    Integer.parseInt(values[2]));
        calculate();
    }
}
/*Contributed and written by Aniket Dey*/

Python3

# Python3 program to find day
# of a given date

def dayofweek(d, m, y):
    t = [ 0, 3, 2, 5, 0, 3,
          5, 1, 4, 6, 2, 4 ]
    y -= m < 3
    return (( y + int(y / 4) - int(y / 100)
             + int(y / 400) + t[m - 1] + d) % 7)

# Driver Code
day = dayofweek(30, 8, 2010)
print(day)

# This code is contributed by Shreyanshi Arun.

C#

// C# program to find day of a given date
using System;

class GFG {

    static int dayofweek(int d, int m, int y)
    {
        int []t = { 0, 3, 2, 5, 0, 3, 5,
                            1, 4, 6, 2, 4 };
        y -= (m < 3) ? 1 : 0;

        return ( y + y/4 - y/100 + y/400
                         + t[m-1] + d) % 7;
    }

    // Driver Program to test above function
    public static void Main()
    {
        int day = dayofweek(30, 8, 2010);

        Console.Write(day);
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find
// day of a given date
function dayofweek($d, $m, $y)
{
    static $t = array(0, 3, 2, 5, 0, 3,
                      5, 1, 4, 6, 2, 4);
    $y -= $m < 3;
    return ($y + $y / 4 - $y / 100 +
            $y / 400 + $t[$m - 1] + $d) % 7;
}

// Driver Code
$day = dayofweek(30, 8, 2010);
echo $day;

// This code is contributed by mits.
?>

Javascript

<script>

// Javascript program to find day of a given date

function dayofweek(d, m, y)
{
    let t = [ 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 ];
    y -= (m < 3) ? 1 : 0;
    return ( y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}

// Driver Code

    let day = dayofweek(30, 8, 2010);
    document.write(Math.round(day));

</script>

Output : 1 (Monday)

Time Complexity: O(1)

Space Complexity: O(1), since no extra space has been taken.

See this for explanation of the above function.
References: 
http://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week
This article is compiled by Dheeraj Jain and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above