#Find the missing letter

Write a method that takes an array of consecutive (increasing) letters as input and that returns the missing letter in the array.

You will always get an valid array. And it will be always exactly one letter be missing. The length of the array will always be at least 2.
The array will always contain letters in only one case.

Example:
```if-not:swift
['a','b','c','d','f'] -> 'e'
['O','Q','R','S'] -> 'P'
```

(Use the English alphabet with 26 letters!)

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Source: codewars | Difficulty: EASY
Tags: Mathematics

Starter code: def find_missing_letter(chars):
	

Solutions:
def find_missing_letter(chars):
	n = 0
	while ord(chars[n]) == ord(chars[n + 1]) - 1:
		n += 1
	return chr(1 + ord(chars[n]))


def find_missing_letter(input):
	alphabet = list('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ')
	start = alphabet.index(input[0])
	for i in range(len(input)):
		if not input[i] == alphabet[start + i]:
			return alphabet[start + i]


def find_missing_letter(c):
	return next((chr(ord(c[i]) + 1) for i in range(len(c) - 1) if ord(c[i]) + 1 != ord(c[i + 1])))


def find_missing_letter(chars):
	for x in range(1, len(chars)):
		if ord(chars[x]) - ord(chars[x - 1]) != 1:
			return chr(ord(chars[x]) - 1)


def find_missing_letter(chars):
	return [chr(n) for n in range(ord(chars[0]), ord(chars[-1]) + 1) if n not in [ord(c) for c in chars]][0]


def find_missing_letter(chars):
	s = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
	match = False
	count = 0
	for letter in s:
		if letter == chars[count]:
			match = True
			count = count + 1
		elif match == True:
			return letter


def find_missing_letter(chars):
	for i in range(0, len(chars) - 1):
		current = chars[i]
		after = chars[i + 1]
		expected = chr(ord(current) + 1)
		if after != expected:
			return expected
	return ''


def find_missing_letter(chars):
	return set((chr(i) for i in range(ord(chars[0]), ord(chars[-1]) + 1))).difference(set(chars)).pop()


def find_missing_letter(chars):
	next = chars[0]
	for letter in chars:
		if letter != next:
			return next
		next = chr(ord(letter) + 1)
	return 0