<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:
        if len(nums) < 2:
            return 0 
        # determine the min and max number in nums array to set out buckets length
        max_num = max(nums)
        min_num = min(nums)

        if max_num == min_num:
            return 0 

        # initialize a buckets array
        bucket_size = max(1, (max_num - min_num) // (len(nums) - 1))
        bucket_count = (max_num - min_num) // bucket_size + 1
        # min and max
        buckets = [[float('inf'), float('-inf')] for _ in range(bucket_count)]

        for num in nums:
            i = (num - min_num) // bucket_size
            buckets[i][0] = min(num, buckets[i][0])
            buckets[i][1] = max(num, buckets[i][1])

        max_value = 0
        previous_max = buckets[0][1]

        for bucket in buckets[1:]:
            if bucket[0] == float('inf'):
                continue
            max_value = max(max_value, bucket[0] - previous_max)
            previous_max = bucket[1]

        return max_value
        



        
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