<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:
        highest, lowest, n = max(nums), min(nums), len(nums)

        if n == 1 or highest == lowest:
            return 0

        buckets = [[] for _ in range(n-1)]
        # ranges are:
        # (high-low)/(n-1)
        bucket_range = (highest-lowest)/(n-1)

        for i, num in enumerate(nums):
            bucket = int((num-lowest)/bucket_range)
            if bucket >= n-1:
                buckets[-1].append(num)
            else:
                buckets[bucket].append(num)

        
        possible = [(min(b), max(b)) for b in buckets if b]
        
        maximum = -1
        for i in range(1, len(possible)):
            maximum = max(maximum, possible[i][0]-possible[i-1][1])
        
        return max(maximum, possible[-1][1]-possible[-1][0])
        

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