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.
- x starts with same day as y.
- 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<iostream> 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 = 2018print(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
<?php // PHP program to find // next identical year // Function for finding extra days // of year more than complete weeks function extraDays($y) { // If current year is a leap year, // then number of weekdays move // ahead by 2 in terms of weekdays. if ($y % 400 == 0 || $y % 100 != 0 && $y % 4 == 0) return 2; // Else number of weekdays // move ahead by 1. return 1; } // Returns next identical year. function nextYear($y) { // Find number of days // moved ahead by y $days = extraDays($y); // Start from next year $x = $y + 1; // Count total number of weekdays // moved ahead so far. for ($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 Code $y = 2018; echo nextYear($y); // This code is contributed by aj_36 ?> |
Javascript
<script> // JavaScript program for the above approach // Function for finding extra days of year // more than complete weeks function 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 || y % 100 != 0 && y % 4 == 0) return 2; // Else number of weekdays move ahead // by 1. return 1; } // Returns next identical year. function nextYear(y) { // Find number of days moved ahead by y let days = extraDays(y); // Start from next year let x = y + 1; // Count total number of weekdays // moved ahead so far. for (let 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 let y = 2018; document.write(nextYear(y)); // This code is contributed by susmitakundugoaldanga. </script> |
Output :
2029
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