<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 Solution:
    def maximumGap(self, nums: List[int]) -> int:
            if len(nums) < 2:
                return 0
            
            min_val, max_val = min(nums), max(nums)
            if min_val == max_val:
                return 0
            
            n = len(nums)
            bucket_size = max(1, (max_val - min_val) // (n - 1))
            bucket_count = (max_val - min_val) // bucket_size + 1
            buckets = [[None, None] for _ in range(bucket_count)]
            
            for num in nums:
                idx = (num - min_val) // bucket_size
                if buckets[idx][0] is None:
                    buckets[idx][0] = buckets[idx][1] = num
                else:
                    buckets[idx][0] = min(buckets[idx][0], num)
                    buckets[idx][1] = max(buckets[idx][1], num)
            
            max_gap = 0
            prev_max = min_val
            
            for bucket in buckets:
                if bucket[0] is not None:
                    max_gap = max(max_gap, bucket[0] - prev_max)
                    prev_max = bucket[1]
                    max_gap = max(max_gap, bucket[1] - bucket[0])
            
            return max_gap
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