# Task
 The year of `2013` is the first year after the old `1987` with only distinct digits.

 Now your task is to solve the following problem: given a `year` number, find the minimum year number which is strictly larger than the given one and has only distinct digits.

# Input/Output


 - `[input]` integer `year`

 `1000 ≤ year ≤ 9000`


 - `[output]` an integer

  the minimum year number that is strictly larger than the input number `year` and all its digits are distinct.

Source: codewars | Difficulty: EASY

Starter code: def distinct_digit_year(year):
	

Solutions:
def distinct_digit_year(year):
	year += 1
	while len(set(str(year))) != 4:
		year += 1
	return year


def distinct_digit_year(year):
	year += 1
	if len(set(str(year))) == len(str(year)):
		return year
	return distinct_digit_year(year)


from itertools import count

def distinct_digit_year(year):
	return next((y for y in count(year + 1) if len(str(y)) == len(set(str(y)))))


def distinct_digit_year(year):
	distinctDigits = False
	while distinctDigits == False:
		year += 1
		if len(set(str(year))) == len(str(year)):
			distinctDigits = True
	return year


def distinct_digit_year(year):
	return next((i for i in range(year + 1, 9999) if len(set(str(i))) == 4))


import re

def distinct_digit_year(year):
	return year + 1 if re.match('(?:([0-9])(?!.*\\1)){4}', str(year + 1)) else distinct_digit_year(year + 1)


def distinct(year):
	year = str(year)
	return len(year) == len(set(year))

def distinct_digit_year(year):
	year += 1
	while not distinct(year):
		year += 1
	return year


distinct_digit_year = lambda y: next((i for i in range(y + 1, 9000) if len(set(str(i))) == 4))


def distinct_digit_year(year):
	next_year = int(year)
	while True:
		sum = 0
		next_year = next_year + 1
		for num in str(next_year):
			sum += str(next_year).count(num)
		if sum == 4:
			break
	return next_year


def distinct_digit_year(year):
	for y in range(year + 1, 9999):
		if len(set(str(y))) == 4:
			return y