<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 deque
import bisect
import math
class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        small = nums[0]
        large = nums[0]
        all_set = set([nums[0]])
        for n in nums[1:]:
            all_set.add(n)
            if n>large:
                large = n
            elif n<small:
                small = n
        if len(all_set) == 1:
            return 0
        bucket_size = math.ceil((large-small)/(len(all_set)-1)) # small and large in two different bucket
        buckets = [None for _ in range(len(all_set))] # (min, max)
        # seen = set()
        for n in all_set:
            # if n in seen:
            #     continue
            idx = (n-small)//bucket_size
            if buckets[idx] is None:
                buckets[idx] = [n, n]
            else:
                buckets[idx][0] = min(buckets[idx][0], n)
                buckets[idx][1] = max(buckets[idx][1], n)
            # seen.add(n)
        x = 0
        while x < len(buckets) and buckets[x] is None:
            x+=1
        max_diff = bucket_size
        pre_max = buckets[x][1]
        for i in range(x+1, len(buckets)):
            if buckets[i] is None:
                continue
            max_diff = max(max_diff, buckets[i][0]-pre_max)
            pre_max = buckets[i][1]
        return max_diff
        
            

        
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