<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:
        minNum = min(nums)
        maxNum = max(nums)
        if len(nums) == 1:
            return 0

        bucketSize = max(1,(maxNum - minNum)//(len(nums)-1))
        print(bucketSize)
        buckets = [[float("inf"), float("-inf")] for _ in range(math.floor((maxNum - minNum)/bucketSize) + 1)]
        print(buckets)
        for num in nums:
            bucket = math.floor((num - minNum)/bucketSize)
            print(bucket)
            buckets[bucket][0] = min(buckets[bucket][0], num)
            buckets[bucket][1] = max(buckets[bucket][1], num)
        
        largestGap = 0
        prevHighest =  buckets[0][1]
        for i in range(1,len(buckets)):
            if buckets[i][0] == float("inf"):
                continue
            largestGap = max(largestGap, buckets[i][0]-prevHighest)
            prevHighest = buckets[i][1]
        
        return largestGap
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