## How to calculate number of leap years between two years in C#

number of leap years between two years c++

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Is there a better way to calculate number of leap years between two years. Assuming I have start date and end date.

I have my code, but I think there should be more elegant way.

calling code:

var numberOfLeapYears = NumberOfLeapYears(startDate.Year + 1, endDate.Year - 1);

function itself:

private static int NumberOfLeapYears(int startYear, int endYear) { var counter = 0; for (var year = startYear; year <= endYear; year++) counter += DateTime.IsLeapYear(year) ? 1 : 0; return counter; }

So if I have `startDate = "10/16/2006"`

and `endDate = "4/18/2004"`

I should only have 1 leap year (2000) in result. Another words startDate's Year and endDate's year should not be calculated, only years in between.

Thanks in advance for your help.

You can count it using analytic approach. A year is a leap year if it can be divided by 4, but can't be divided by 100, except of case when it can be divided by 400. Assuming that you can count such number by following code:

static int LeapYearsBetween(int start, int end) { System.Diagnostics.Debug.Assert(start < end); return LeapYearsBefore(end) - LeapYearsBefore(start + 1); } static int LeapYearsBefore(int year) { System.Diagnostics.Debug.Assert(year > 0); year--; return (year / 4) - (year / 100) + (year / 400); }

Some kind of math magic. It is much effective solution than using LINQ.

**C# Sharp Exercises: Find the leap years between to specified years ,** C# Sharp programming, exercises, solution: Write a C# Sharp program to find the leap years between Write a C# Sharp program to find the leap years between 1994 and 2016. One year from 2/29/1996 is 2/28/1997. Previous: Write a C# Sharp program to determine the type of a particular object. In this post, we are going to write a C program to print leap years between two given years. So basically we need to collect the start year and end year from the user, then we need to loop between the start and end year and then find the leap years to print. But before if you want to know the formula to find the leap year here it is, 1.

You can do it with LINQ simply as bellow:

var leepYears = Enumerable.Range(startYear, endYear - startYear + 1) .Count(x => DateTime.IsLeapYear(x));

**How to calculate number of leap days between two dates.?,** I know how to calculate between year but m confused with dates. I am assuming that you are looking for no of leap years between two date if the date2 is crossing the 02/29/16 then only it give the result "Leap Year" If Date1 is in A1 and Date2 in A2 please try: =IF(AND(A1<DATE(2016,2,29),A2>DATE(2016,2,29)),"Leap Year","No Leap Year")

This should perform much better for large spans of time:

public int LeapYearsBetween(int year1, int year2) { var y1 = new DateTime(year1, 1, 1); var y2 = new DateTime(year2, 1, 1); var nonLeapDays = 365 * (y2.Year - y1.Year); var leapDays = (y2 - y1).Days - nonLeapDays; return leapDays; }

Note that this counts the earlier year if it is a leap year, but not the later year. You'll need to modify the function if you need different behavior.

**Finding the number of leap years between two dates in C# – http ,** I recently ran into a situation where I needed to figure out if there were any leap years between two dates. Although I'm quite sure there is a� Years are abstractions though, so unless you try to deliberately bracket February 29 in a leap year you might not even know a leap year is in the calculation. However, leap years are in the tool. If you see a day difference that looks “off by one”, it’s actually correct and you’re seeing the extra leap year day counted.

Another Linq :-)

int start = 1980; int end = 2000; var count = Enumerable.Range(start, end - start + 1) .Aggregate(0, (a, b) => DateTime.IsLeapYear(b) ? a + 1 : a);

**Method to determine whether a year is a leap year,** To determine whether a year is a leap year, follow these steps: If the year is evenly divisible by 4, go to step 2. If the year is evenly divisible by 100, go to step 3. If the year is evenly divisible by 400, go to step 4. The year is a leap year (it has 366 days). The year is not a leap year (it has 365 days). To calculate whether or not a year is a leap year, start by checking if the number is evenly divisible by 4 since leap years need to be. Then, if it is evenly divisible by 4, check to see if it's evenly divisible by 100. If it's not, then it's a leap year. However, if it is evenly divisible by both 100 and 4, it might still be a leap year.

Just another code with primitives types

public static int CountLeapYears(int startYear, int endYear) { int acc = 0; while (true) { if ((startYear % 4 == 0 && startYear % 100 != 0) || startYear % 400 == 0) acc++; if (startYear + 1 == endYear) break; startYear++; } return acc; }

**DateTime.IsLeapYear(Int32) Method (System),** Returns an indication whether the specified year is a leap year. method to determine which years between 1994 and 2014 are leap years. The example also illustrates the result when the AddYears method is used to add a year to a leap day. C# Copy year is specified as a 4-digit base 10 number; for example, 1996. The number of leap years between two dates in England can be different than the number of leap years between two dates in the US for example. – Eric Lippert Jan 3 '11 at 23:18 @Eric Lippert.

**Count of Leap Years in a given year range,** Given two years L and R, the task is to find the total number of leap years So calculate the numbers which satisfy above condition in range (1, L) and (1, R) by. Home Tutorials Java Corejava Datetimes Calculate number of leap year between to years in java. Calculate number of leap year between to years in java. Posted in : Core Java Posted on : December 4, 2010 at 5:55 PM Comments : [ 0 ]

**Leap Years List - List of Leap Years Between Two Dates,** This tool is used to list all leap years between two years. Leap years. Leap years are those years divisible by 4, except for century years whose number is not� Based on my understanding, if you want to get the number of leap days between two dates, you could refer to following method in you data base. select datediff(dd,'2014-8-1','2014-9-3') More information about how to calculate number of leap days between two dates, please refer to the link below:

**Calculate Age Between Two Dates,** The reason we need to handle the leap years is because in a leap year there are 366 days vs Calculate and return years between two dates. This program will read value of N and print all Leap Years from 1 to N years.There are two conditions for leap year: 1-If year is divisible by 400 ( for Century years), 2-If year is divisible by 4 and must not be divisible by 100 (for Non Century years).

##### Comments

- This seems like a reasonable implementation
- I assume you meant 10/16/1996. In this case, you don't care about the 2/29/2004, which was a leap day that occurred prior to the end date of 4/18/2004?
- Do you need to take into account that the leap year calculations were different before the Gregorian calendar reform? Also, do you need to take into account that some countries adopted the reformed calendar at different times? The number of leap years between two dates
*in England*can be different than the number of leap years between two dates*in the US*for example. - @Eric Lippert. No I don't need to take into account leap years different before Gregorian calendar reform. I don't need to take into account any countris (US only). Thanks Eric.
- If you think of the number of loops, you can accelerate the process starting jumping by 4 years once you find the first one.
- I like this approach since there is no iteration for each year. In the begging I like more LINQ approach, but after analyzing your code I found it should not have impact on performance since LINQ approach need to go through each year and find if it is leap or not. Yours solution just pure math. I tested and it gives the same result as LINQ. Thanks again.
- @Vlad Bezden, currently all the years are bellow 10000, so there is no feasible performance issue and because it will be done in memory it's not bad in all, in this cases I personally prefer clean code, if it was for database read write, yes it would be better to do it better.
- Personally, I'd try to stick to O(1) algorithms over O(n) wherever possible.
- FYI: Last method is the same as "(year-1) * 0.2425" if using doubles. :)
- This is pretty fantastic! Thanks so much for sharing.
- @Vlad Bezden LINQ is great tool, but for this task there is much efficient solution (see my answer). When range is large LINQ solution is slow, but my algorithm is independent from range size.
- +1 I really this code - it's easy to read and the intent of code is very clear.
- @Saeed yes I like how code is clean, but agree with mace that LINQ solution is slower especially when we have big difference between startYear and endYear.