<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):
        self.used = False
        self.min_val = float("inf")
        self.max_val = -float("inf")

class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        if len(nums) < 2:
            return 0
        
        min_val, max_val = min(nums), max(nums)
        bucket_size = max(1, (max_val - min_val) // (len(nums) - 1) - 1)
        bucket_num = (max_val - min_val) // bucket_size + 1
        buckets = [Bucket() for _ in range(bucket_num)]

        for num in nums:
            idx = (num - min_val) // bucket_size
            buckets[idx].used = True
            buckets[idx].min_val = min(buckets[idx].min_val, num)
            buckets[idx].max_val = max(buckets[idx].max_val, num)
        
        max_gap = 0
        prev_bucket_max = min_val
        for bucket in buckets:
            if not bucket.used: 
                continue
            max_gap = max(max_gap, bucket.min_val - prev_bucket_max)
            prev_bucket_max = bucket.max_val

        return max_gap


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