<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:
        nums = list(set(nums))
        if len(nums) < 2:
            return 0
        mn = min(nums)
        mx = max(nums)

        bucket_count = len(nums)
        bucket_size = (mx - mn) // (len(nums)) + 1
        if bucket_size < 2:
            return 1
        maxs = [0 for _ in range(bucket_count)]
        mins = [sys.maxsize for _ in range(bucket_count)]
        for num in nums:
            bucket_number = self.__bucket_number(num, mn, bucket_size)
            maxs[bucket_number] = max(maxs[bucket_number], num)
            mins[bucket_number] = min(mins[bucket_number], num)
        prev_max = maxs[0]
        result = 0
        for mn, mx in zip(mins, maxs):
            if mx > 0:
                result = max(result, mn - prev_max)
                prev_max = mx
        return result


    def __bucket_number(self, num, mn, bucket_size):
        return (num - mn) // bucket_size
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