# Find the next identical calendar year

You are given an year Y, find the next identical calendar year to Y.
Examples :

```Input : 2017
Output : 2023

Input : 2018
Output : 2029```

An year x is identical to a given previous year y if following two conditions are satisfied.

1. x starts with same day as y.
2. If y is leap year, then x is also. If y is not leap year, then x is also not.

The idea is to check all years one by one (starting from next year). We keep track of number of days moved ahead. If total moved days is 7, then current year begins with same day. We also check if leap-ness of current year is same as y. If both conditions are satisfied, we return current year.

## C++

 `// C++ program to find next identical year ``#include ``using` `namespace` `std; `` ` `// Function for finding extra days of year ``// more than complete weeks ``int` `extraDays(``int` `y) ``{ ``    ``// If current year is a leap year, then ``    ``// it number of weekdays move ahead by ``    ``// 2 in terms of weekdays. ``    ``if` `(y%400==0 || y%100!=0 && y%4==0) ``        ``return` `2; `` ` `    ``// Else number of weekdays move ahead ``    ``// by 1. ``    ``return` `1; ``} `` ` `// Returns next identical year. ``int` `nextYear(``int` `y) ``{ ``    ``// Find number of days moved ahead by y ``    ``int` `days = extraDays(y); `` ` `    ``// Start from next year ``    ``int` `x = y + 1; `` ` `    ``// Count total number of weekdays ``    ``// moved ahead so far. ``    ``for` `(``int` `sum=0; ; x++) ``    ``{ ``        ``sum = (sum + extraDays(x)) % 7; `` ` `        ``// If sum is divisible by 7 and leap-ness ``        ``// of x is same as y, return x. ``        ``if` `( sum==0 && (extraDays(x) == days)) ``            ``return` `x; ``    ``} `` ` `    ``return` `x; ``} `` ` `// driver program ``int` `main() ``{ ``    ``int` `y = 2018; ``    ``cout << nextYear(y); ``    ``return` `0; ``} `

## Java

 `// Java program to find next identical year ``class` `GFG { `` ` `// Function for finding extra days of year ``// more than complete weeks ``static` `int` `extraDays(``int` `y) ``{ ``    ``// If current year is a leap year, then ``    ``// it number of weekdays move ahead by ``    ``// 2 in terms of weekdays. ``    ``if` `(y % ``400` `== ``0` `|| y % ``100` `!= ``0` `&& y % ``4` `== ``0``) ``        ``return` `2``; `` ` `    ``// Else number of weekdays move ahead ``    ``// by 1. ``    ``return` `1``; ``} `` ` `// Returns next identical year. ``static` `int` `nextYear(``int` `y) ``{ ``    ``// Find number of days moved ahead by y ``    ``int` `days = extraDays(y); `` ` `    ``// Start from next year ``    ``int` `x = y + ``1``; `` ` `    ``// Count total number of weekdays ``    ``// moved ahead so far. ``    ``for` `(``int` `sum = ``0``; ; x++) ``    ``{ ``        ``sum = (sum + extraDays(x)) % ``7``; `` ` `        ``// If sum is divisible by 7 and leap-ness ``        ``// of x is same as y, return x. ``        ``if` `( sum == ``0` `&& (extraDays(x) == days)) ``            ``return` `x; ``    ``} `` ` `} `` ` `// Driver code ``public` `static` `void` `main(String[] args) ``{ ``    ``int` `y = ``2018``; ``    ``System.out.println(nextYear(y)); ``} ``} `` ` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to find next identical year `` ` `# Function for finding extra days of year ``# more than complete weeks ``def` `extraDays(y) : `` ` `    ``# If current year is a leap year, then ``    ``# it number of weekdays move ahead by ``    ``# 2 in terms of weekdays. ``    ``if` `(y ``%` `400` `=``=` `0` `or` `y ``%` `100` `!``=` `0` `and` `y ``%` `4` `=``=` `0``) : ``        ``return` `2`` ` `    ``# Else number of weekdays move ahead ``    ``# by 1. ``    ``return` `1`` ` `# Returns next identical year. ``def` `nextYear(y) : `` ` `    ``# Find number of days moved ahead by y ``    ``days ``=` `extraDays(y) `` ` `    ``# Start from next year ``    ``x ``=` `y ``+` `1`` ` `    ``# Count total number of weekdays ``    ``# moved ahead so far. ``    ``Sum` `=` `0``    ``while``(``True``) : `` ` `        ``Sum` `=` `(``Sum` `+` `extraDays(x)) ``%` `7`` ` `        ``# If sum is divisible by 7 and leap-ness ``        ``# of x is same as y, return x. ``        ``if` `( ``Sum` `=``=` `0` `and` `(extraDays(x) ``=``=` `days)) : ``            ``return` `x ``             ` `        ``x ``+``=` `1`` ` `    ``return` `x `` ` `y ``=` `2018``print``(nextYear(y)) `` ` `# This code is contributed by mukesh07.`

## C#

 `// C# program to find next identical year ``using` `System; ``     ` `class` `GFG  ``{ `` ` `// Function for finding extra days of year ``// more than complete weeks ``static` `int` `extraDays(``int` `y) ``{ ``    ``// If current year is a leap year, then ``    ``// it number of weekdays move ahead by ``    ``// 2 in terms of weekdays. ``    ``if` `(y % 400 == 0 || y % 100 != 0 && y % 4 == 0) ``        ``return` `2; `` ` `    ``// Else number of weekdays move ahead ``    ``// by 1. ``    ``return` `1; ``} `` ` `// Returns next identical year. ``static` `int` `nextYear(``int` `y) ``{ ``    ``// Find number of days moved ahead by y ``    ``int` `days = extraDays(y); `` ` `    ``// Start from next year ``    ``int` `x = y + 1; `` ` `    ``// Count total number of weekdays ``    ``// moved ahead so far. ``    ``for` `(``int` `sum = 0; ; x++) ``    ``{ ``        ``sum = (sum + extraDays(x)) % 7; `` ` `        ``// If sum is divisible by 7 and leap-ness ``        ``// of x is same as y, return x. ``        ``if` `( sum == 0 && (extraDays(x) == days)) ``            ``return` `x; ``    ``} `` ` `} `` ` `// Driver code ``public` `static` `void` `Main(String[] args) ``{ ``    ``int` `y = 2018; ``    ``Console.WriteLine(nextYear(y)); ``} ``} `` ` `// This code has been contributed by 29AjayKumar `

## PHP

 ` `

## Javascript

 ` `

Output :

`2029`

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