An ordered sequence of numbers from 1 to N is given. One number might have deleted from it, then the remaining numbers were mixed. Find the number that was deleted.

Example: 
 - The starting array sequence is `[1,2,3,4,5,6,7,8,9]`
 - The mixed array with one deleted number is `[3,2,4,6,7,8,1,9]`
 - Your function should return the int `5`.

If no number was deleted from the array and no difference with it, your function should return the int `0`.

Note that N may be 1 or less (in the latter case, the first array will be `[]`).

Source: codewars | Difficulty: EASY
Tags: Data structures

Starter code: def find_deleted_number(arr, mixed_arr):
	

Solutions:
def find_deleted_number(arr, mixed_arr):
	return sum(arr) - sum(mixed_arr)


def find_deleted_number(a, b):
	return (set(a) - set(b)).pop() if len(a) != len(b) else 0


def find_deleted_number(arr, mixed_arr):
	lost = 0
	for i in arr:
		if i not in mixed_arr:
			lost = i
	return lost


find_deleted_number = lambda a, m: next((e for e in set(a) - set(m)), 0)


def find_deleted_number(arr, mixed_arr):
	return 0 if not arr else arr[-1] * (arr[-1] + 1) / 2 - sum(mixed_arr)


def find_deleted_number(arr, mixed_arr):
	return len(arr) > len(mixed_arr) and (set(arr) - set(mixed_arr)).pop()


def find_deleted_number(arr, mixed_arr):
	res = []
	for n in arr:
		if n not in mixed_arr:
			return n
	return 0


def find_deleted_number(arr, mixed_arr):
	ret = 0
	for i in arr:
		if i not in mixed_arr:
			ret = i
	return ret


def find_deleted_number(arr, mixed_arr):
	arr = set(sorted(arr))
	mixed_arr = set(sorted(mixed_arr))
	return 0 if arr == mixed_arr else list(arr.difference(mixed_arr))[0]