<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:
        n = len(nums)
        if n < 2 : return 0
        mini = min(nums)
        maxi = max(nums)
        diff = maxi - mini
        bucket_size = max(1, diff / (n - 1))
        bucket = defaultdict(list)

        for num in nums:
            key = (num - mini) // bucket_size
            if key not in bucket:
                bucket[key] = [num, num]
            else:
                bucket[key] = [max(bucket[key][0], num), min(bucket[key][1], num)]

        res = 0
        preKey = -1
        for key in sorted(bucket.keys()):
            if preKey != -1:
                res = max(res, bucket[key][1] - bucket[preKey][0])
            preKey = key

        return res
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