<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:
        n = len(nums)
        if n < 2: return 0

        def countsort(nums, place):
            K = 10
            counts = [0] * K

            for el in nums:
                i = (el // place) % K
                counts[i] += 1
            
            idx = 0
            for i, count in enumerate(counts):
                counts[i] = idx
                idx += count

            sortedlist = [0] * n
            for el in nums:
                i = (el // place) % 10
                sortedlist[counts[i]] = el
                counts[i] += 1
        
            nums[:] = [num for num in sortedlist]



        def radix_sort(nums):
            shift = min(nums)
            nums[:] = [num - shift for num in nums]
            maxel = max(nums)

            place = 1
            while place <= maxel:
                countsort(nums, place)
                place *= 10
            nums[:] = [num + shift for num in nums]

        heapify(nums)
        maxGap = 0
        last = heappop(nums)
        while nums:
            maxGap = max(maxGap, nums[0] - last)
            last = heappop(nums)
        return maxGap

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