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


time.perf_counter() always returns the float value of time in seconds. while pref_counter_ns() always gives the integer value of time in nanoseconds.


t1_start = perf_counter()
t1_stop = perf_counter()
print("Elapsed time:", t1_stop, t1_start)