<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
    
        minVal = min(nums)
        maxVal = max(nums)
        
        if minVal == maxVal:
            return 0
        
        n = len(nums)
        bucketSize = max((maxVal - minVal) // (n - 1), 1)
        bucketCount = (maxVal - minVal) // bucketSize + 1
        
        # Initialize buckets
        buckets = [[float('inf'), float('-inf')] for _ in range(bucketCount)]
        
        # Place each number in a bucket
        for num in nums:
            idx = (num - minVal) // bucketSize
            buckets[idx][0] = min(buckets[idx][0], num)
            buckets[idx][1] = max(buckets[idx][1], num)
        
        # Find the maximum gap
        maxGap = 0
        prevMax = buckets[0][1]
        
        for i in range(1, bucketCount):
            if buckets[i][0] == float('inf'):
                continue
            maxGap = max(maxGap, buckets[i][0] - prevMax)
            prevMax = buckets[i][1]
        
        return maxGap
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