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

<p>You must write an algorithm that runs in linear time and uses linear extra space.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,6,9,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Bucket:
  def __init__(self, mn: int, mx: int):
    self.mn = mn
    self.mx = mx


class Solution:
  def maximumGap(self, nums: list[int]) -> int:
    if len(nums) < 2:
      return 0

    mn = min(nums)
    mx = max(nums)
    if mn == mx:
      return 0

    gap = math.ceil((mx - mn) / (len(nums) - 1))
    bucketSize = (mx - mn) // gap + 1
    buckets = [Bucket(math.inf, -math.inf) for _ in range(bucketSize)]

    for num in nums:
      i = (num - mn) // gap
      buckets[i].mn = min(buckets[i].mn, num)
      buckets[i].mx = max(buckets[i].mx, num)

    ans = 0
    prevMax = mn

    for bucket in buckets:
      if bucket.mn == math.inf:
        continue  # empty bucket
      ans = max(ans, bucket.mn - prevMax)
      prevMax = bucket.mx

    return ans
37071