Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Return 0 if the array contains less than 2 elements.
Example 1:
Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
             (3,6) or (6,9) has the maximum difference 3.
Example 2:
Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.
Note:
       You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
       Try to solve it in linear time/space.
class Solution:
     def maximumGap(self, num):
         if len(num) < 2 or min(num) == max(num):
             return 0
         a, b = min(num), max(num)
         import math
         size = math.ceil((b - a) / (len(num) - 1))
         bucket = [[None, None] for _ in range((b - a) // size + 1)]
         for n in num:
             b = bucket[(n - a) // size]
             b[0] = n if b[0] is None else min(b[0], n)
             b[1] = n if b[1] is None else max(b[1], n)
         bucket = [b for b in bucket if b[0] is not None]
         return max(bucket[i][0] - bucket[i - 1][1] for i in range(1, len(bucket)))
class Solution:
     def maximumGap(self, num):
         if len(num) < 2 or min(num) == max(num):
             return 0
         a, b = min(num), max(num)
         import math
         size = math.ceil((b - a) / (len(num) - 1))
         bucket = [[None, None] for _ in range((b - a) // size + 1)]
         for n in num:
             b = bucket[(n - a) // size]
             b[0] = n if b[0] is None else min(b[0], n)
             b[1] = n if b[1] is None else max(b[1], n)
         bucket = [b for b in bucket if b[0] is not None]
         return max(bucket[i][0] - bucket[i - 1][1] for i in range(1, len(bucket)))