# Task
 Given an integer `n`, find the maximal number you can obtain by deleting exactly one digit of the given number.

# Example

 For `n = 152`, the output should be `52`;
 
 For `n = 1001`, the output should be `101`.
 
# Input/Output


 - `[input]` integer `n`

    Constraints: `10 ≤ n ≤ 1000000.`


 - `[output]` an integer

Source: codewars | Difficulty: EASY
Tags: Ad-hoc

Starter code: def delete_digit(n):
	

Solutions:
def delete_digit(n):
	s = str(n)
	return int(max((s[:i] + s[i + 1:] for i in range(len(s)))))


def delete_digit(n):
	liststrn = list(str(n))
	for i in range(len(liststrn)):
		try:
			if liststrn[i + 1] > liststrn[i]:
				return int(''.join(liststrn[:i] + liststrn[i + 1:]))
		except IndexError:
			return int(''.join(liststrn[:-1]))


def delete_digit(n):
	for z in range(len(str(n)) - 1):
		if int(str(n)[z]) < int(str(n)[z + 1]):
			return int(str(n)[:z] + str(n)[z + 1:])
	return int(str(n)[:-1])


def delete_digit(n):
	n = str(n)
	res = 0
	for i in range(len(n)):
		digit = int(n[0:i] + n[i + 1:])
		if digit > res:
			res = digit
	return res


from itertools import combinations

def delete_digit(n):
	return max((int(''.join(i)) for i in list(combinations(str(n), len(str(n)) - 1))))


def delete_digit(n):
	max = 0
	str_n = str(n)
	for i in range(len(str_n)):
		temp = str_n[:i] + str_n[i + 1:]
		if max < int(temp):
			max = int(temp)
	return max


def delete_digit(n):
	(n, i) = (str(n), 0)
	while i < len(n) - 1 and n[i] >= n[i + 1]:
		i += 1
	return int(n[:i] + n[i + 1:])


def delete_digit(n):
	return max([int(''.join([ch for (index, ch) in enumerate(str(n)) if index != str(n).index(i)])) for i in str(n)])


def delete_digit(n):
	numbers = []
	nl = list(str(n))
	for i in range(len(nl)):
		del nl[i]
		number = int(''.join(nl))
		numbers.append(number)
		nl = list(str(n))
	numbers.sort()
	numbers = numbers[::-1]
	return numbers[0]