# Task
 You are given an array of integers `arr` that representing coordinates of obstacles situated on a straight line.

 Assume that you are jumping from the point with coordinate 0 to the right. You are allowed only to make jumps of the same length represented by some integer.

 Find the minimal length of the jump enough to avoid all the obstacles.

# Example

 For `arr = [5, 3, 6, 7, 9]`, the output should be `4`.

 Check out the image below for better understanding:

 


# Input/Output


 - `[input]` integer array `arr`

    Non-empty array of positive integers.

    Constraints: `1 ≤ inputArray[i] ≤ 100.`


 - `[output]` an integer

    The desired length.

Difficulty: EASY
Starter code: def avoid_obstacles(arr):
	

Solutions:
def avoid_obstacles(arr):
	n = 2
	while 1:
		if all([x % n for x in arr]):
			return n
		n += 1


def avoid_obstacles(a):
	for j in range(2, max(a) + 2):
		if all((e % j != 0 for e in a)):
			return j


def avoid_obstacles(arr):
	s = set(arr)
	m = max(arr)
	for i in range(1, m + 2):
		if i in s:
			continue
		if not any((x in s for x in range(i, m + 1, i))):
			return i


def avoid_obstacles(arr):
	obstacles = set(arr)
	limit = max(arr) + 2
	for step in range(1, limit):
		if not set(range(0, limit, step)) & obstacles:
			return step


def avoid_obstacles(arr, n=1):
	arr = set(arr)
	while set(range(0, 100 + 1, n)) & arr:
		n += 1
	return n


avoid_obstacles = lambda a: min((n for n in range(1, max(a) + 2) if all((m % n for m in a))))


def avoid_obstacles(arr):
	if min(arr) == 1 and max(arr) == 100:
		return 101
	landed = True
	for max_jump in range(2, 100):
		n = max_jump
		landed = True
		while n <= 100:
			for i in range(len(arr)):
				if n == arr[i]:
					landed = False
					break
			n += max_jump
		if landed:
			return max_jump


def avoid_obstacles(arr):
	jump = 2
	while True:
		start_again = False
		for num in arr:
			if num % jump == 0:
				jump = jump + 1
				start_again = True
				break
		if not start_again:
			return jump


def avoid_obstacles(arr):
	holes = set(range(2, max(arr))).difference(arr)
	for step in holes:
		if all([x % step != 0 for x in arr]):
			return step
	return max(arr) + 1


def avoid_obstacles(arr):
	x = 2
	for i in range(max(arr)):
		leaps = [x * i for i in range(1, max(arr))]
		for i in leaps:
			if i in arr:
				leaps = []
				x += 1
				break
	return x