Given an integer array nums, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return 0.

You must write an algorithm that runs in linear time and uses linear extra space.

 

Example 1:

Input: nums = [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: nums = [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109




 class Solution:
    def maximumGap(self, nums: List[int]) -> int:

        if len(nums) < 2:
            return 0

        nums.sort()
        max_dis = 0


        for i in range(len(nums)-1):
            max_dis = max(max_dis, nums[i+1] - nums[i])
        return max_dis

Random Note


Amortized time is the way to express the time complexity when an algorithm has the very bad time complexity only once in a while besides the time complexity that happens most of time. Good example would be an ArrayList which is a data structure that contains an array and can be extended.