Find the first day of a given year from a base year having first day as Monday

Given two integers Y and B representing two years, the task is to find the day of the week in which 1^st January of the year Y lies assuming that 1^st January of the year B was a Monday.

Examples:

Input:
Y = 2020
B = 1900
Output:
Wednesday
Explanation:
01/01/2020 was a Wednesday considering that 01/01/1900 was a Monday

Input:
Y = 2020
B = 1905
Output:
Thursday
Explanation:
01/01/2020 was a Wednesday assuming that 01/01/1905 was a Monday

Approach: Follow the steps below to solve the problem:

1. Total years lying between base year (B) and the year (Y) is equal to (Y – 1) – B.
2. Total number of leap years lying in between = Total years / 4
3. Total number of non-leap years in between = Total Years – Leap Years.
4. Total Days = Total Leap Years * 366 + Non-Leap Years * 365 + 1.
5. Therefore, the day of the 1^st January of Y is Total Days % 7.

Below is the implementation of the above approach:

C++14

// C++ Implementation of
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the day of
// 1st January of Y year
void findDay(int Y, int B)
{
    int lyear, rest, totaldays, day;

    // Count years between
    // years Y and B
    Y = (Y - 1) - B;

    // Count leap years
    lyear = Y / 4;

    // Non leap years
    rest = Y - lyear;

    // Total number of days in the years
    // lying between the years Y and B
    totaldays = (rest * 365)
                + (lyear * 366) + 1;

    // Actual day
    day = (totaldays % 7);

    if (day == 0)
        printf("Monday");

    else if (day == 1)
        printf("Tuesday");

    else if (day == 2)
        printf("Wednesday");

    else if (day == 3)
        printf("Thursday");

    else if (day == 4)
        printf("Friday");

    else if (day == 5)
        printf("Saturday");

    else if (day == 6)
        printf("Sunday");

    else
        printf("INPUT YEAR IS WRONG!");
}

// Driver Code
int main()
{
    int Y = 2020, B = 1900;
    findDay(Y, B);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG
{

  // Function to find the day of
  // 1st January of Y year
  static void findDay(int Y, int B)
  {
    int lyear, rest, totaldays, day;

    // Count years between
    // years Y and B
    Y = (Y - 1) - B;

    // Count leap years
    lyear = Y / 4;

    // Non leap years
    rest = Y - lyear;

    // Total number of days in the years
    // lying between the years Y and B
    totaldays = (rest * 365)
      + (lyear * 366) + 1;

    // Actual day
    day = (totaldays % 7);

    if (day == 0)
      System.out.println("Monday");

    else if (day == 1)
      System.out.println("Tuesday");

    else if (day == 2)
      System.out.println("Wednesday");

    else if (day == 3)
      System.out.println("Thursday");

    else if (day == 4)
      System.out.println("Friday");

    else if (day == 5)
      System.out.println("Saturday");

    else if (day == 6)
      System.out.println("Sunday");

    else
      System.out.println("INPUT YEAR IS WRONG!");
  }

  // Driver code
  public static void main(String[] args)
  {
    int Y = 2020, B = 1900;
    findDay(Y, B);
  }
}

// This code is contributed by code_hunt.

Python3

# Python program to implement
# the above approach

# Function to find the day of
# 1st January of Y year
def findDay(Y, B):
    lyear, rest, totaldays, day = 0, 0, 0, 0;

    # Count years between
    # years Y and B
    Y = (Y - 1) - B;

    # Count leap years
    lyear = Y // 4;

    # Non leap years
    rest = Y - lyear;

    # Total number of days in the years
    # lying between the years Y and B
    totaldays = (rest * 365) + (lyear * 366) + 1;

    # Actual day
    day = (totaldays % 7);

    if (day == 0):
        print("Monday");

    elif (day == 1):
        print("Tuesday");

    elif (day == 2):
        print("Wednesday");

    elif (day == 3):
        print("Thursday");

    elif (day == 4):
        print("Friday");

    elif (day == 5):
        print("Saturday");

    elif (day == 6):
        print("Sunday");

    else:
        print("INPUT YEAR IS WRONG!");

# Driver code
if __name__ == '__main__':
    Y = 2020; B = 1900;
    findDay(Y, B);

# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
class GFG
{
  // Function to find the day of
  // 1st January of Y year
  static void findDay(int Y, int B)
  {
    int lyear, rest, totaldays, day;

    // Count years between
    // years Y and B
    Y = (Y - 1) - B;

    // Count leap years
    lyear = Y / 4;

    // Non leap years
    rest = Y - lyear;

    // Total number of days in the years
    // lying between the years Y and B
    totaldays = (rest * 365)
      + (lyear * 366) + 1;

    // Actual day
    day = (totaldays % 7);
    if (day == 0)
      Console.WriteLine("Monday");
    else if (day == 1)
      Console.WriteLine("Tuesday");
    else if (day == 2)
      Console.WriteLine("Wednesday");
    else if (day == 3)
      Console.WriteLine("Thursday");
    else if (day == 4)
      Console.WriteLine("Friday");
    else if (day == 5)
      Console.WriteLine("Saturday");
    else if (day == 6)
      Console.WriteLine("Sunday");
    else
      Console.WriteLine("INPUT YEAR IS WRONG!");
  }

// Driver code
static void Main()
{
    int Y = 2020, B = 1900;
    findDay(Y, B);
}
}

// This code is contribute by susmitakundugoaldanga

Javascript

<script>

// Javascript program of the above approach

  // Function to find the day of
  // 1st January of Y year
  function findDay(Y, B)
  {
    let lyear, rest, totaldays, day;

    // Count years between
    // years Y and B
    Y = (Y - 1) - B;

    // Count leap years
    lyear = Math.floor(Y / 4);

    // Non leap years
    rest = Y - lyear;

    // Total number of days in the years
    // lying between the years Y and B
    totaldays = (rest * 365)
      + (lyear * 366) + 1;

    // Actual day
    day = (totaldays % 7);

    if (day == 0)
      document.write("Monday");

    else if (day == 1)
      document.write("Tuesday");

    else if (day == 2)
      document.write("Wednesday");

    else if (day == 3)
      document.write("Thursday");

    else if (day == 4)
      document.write("Friday");

    else if (day == 5)
      document.write("Saturday");

    else if (day == 6)
      document.write("Sunday");

    else
      document.write("INPUT YEAR IS WRONG!");
  }

    // Driver Code

    // Given array
    let Y = 2020, B = 1900;
    findDay(Y, B);

</script>

Output:

Wednesday

Time Complexity: O(1)
Auxiliary Space: O(1)