# Python | Print number of leap years from given list of years

The problem of finding leap year is quite generic and we might face the issue of finding the number of leap years in given list of years. Let’s discuss certain ways in which this can be performed.

Method #1: Using Iteration

Check whether year is a multiple of 4 and not multiple of 100 or year is multiple of 400.

 `# Python code to finding number of  ` `# leap years in list of years. ` ` `  `# Input list initialization ` `Input` `=` `[``2001``, ``2002``, ``2003``, ``2004``, ``2005``, ``2006``, ` `         ``2007``, ``2008``, ``2009``, ``2010``, ``2011``, ``2012``] ` ` `  `# Find whether it is leap year or not ` `def` `checkYear(year): ` `    ``return` `(((year ``%` `4` `=``=` `0``) ``and` `             ``(year ``%` `100` `!``=` `0``)) ``or` `             ``(year ``%` `400` `=``=` `0``))  ` `  `  `# Answer Initialization ` `Answer ``=` `0`  ` `  `for` `elem ``in` `Input``: ` `    ``if` `checkYear(elem): ` `        ``Answer ``=` `Answer ``+` `1` ` `  `# Printing ` `print``(``"No of leap years are:"``, Answer) `

Output:

```No of leap years are: 3
```

Method #2: Using `calender`

 `# Python code to finding number of  ` `# leap years in list of years. ` ` `  `# Importing calender ` `import` `calendar ` ` `  `# Input list initialization ` `Input` `=` `[``2001``, ``2002``, ``2003``, ``2004``, ``2005``, ` `         ``2006``, ``2007``, ``2008``, ``2009``, ``2010``] ` ` `  `# Using calender to find leap year ` `def` `FindLeapYear(``Input``): ` `    ``ans ``=` `0` `    ``for` `elem ``in` `Input``: ` `        ``if` `calendar.isleap(``int``(elem)): ` `            ``ans ``=` `ans ``+` `1` `    ``return` `ans ` ` `  `Output ``=` `FindLeapYear(``Input``) ` ` `  `# Printing ` `print``(``"No of leap years are:"``, Output) `

Output:

```No of leap years are: 2
```

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