<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>

from collections import defaultdict
class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        if len(nums) == 2:
            return abs(nums[1] - nums[0])
        max_num, min_num = max(nums), min(nums)
        bucket = [[float('-inf'), float('inf')] for _ in range(len(nums) - 1)]
        for num in nums:
            index = (num - min_num) * (len(nums) - 1) // (max_num - min_num) if num != max_num else len(nums) - 2
            bucket[index][0] = max(bucket[index][0], num)
            bucket[index][1] = min(bucket[index][1], num)
        result = 0
        prev = 0
        for i in range(1, len(bucket)):
            if bucket[i][1] == float('inf'):
                continue
            while bucket[prev][0] == float('-inf'):
                prev += 1
                if prev >= len(bucket) - 1:
                    return result
            result = max(result, bucket[i][1] - bucket[prev][0])
            prev += 1
        print(bucket)
        return result
42881