<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 bucketSort(self, nums: List[int]) -> List[int]:
        numBuckets = len(nums)
        maxVal = max(nums)
        minVal = min(nums)

        if minVal == maxVal:
            return nums

        bucketRange = (maxVal - minVal) / numBuckets
        buckets = [[] for _ in range(numBuckets)]

        for num in nums:
            bucketIndex = int((num - minVal) / bucketRange)
            # Handle the edge case where num == max_value
            if bucketIndex == numBuckets:
                bucketIndex -= 1
            buckets[bucketIndex].append(num)
        
        buckets = [sorted(bucket) for bucket in buckets]

        sortedArr = []
        for bucket in buckets:
            sortedArr += bucket
        return sortedArr
    

    def maximumGap(self, nums: List[int]) -> int:
        # for linear time algo, which cant be nlogn, use bubble sort, closes to o(n)

        # The average complexity of Bucket Sort is O(n + k log k), which can be very close to O(n) if 
        # the elements are uniformly distributed across buckets and each bucket contains only a few elements. 
        # This makes Bucket Sort faster in practice under the right conditions, but it isn't always O(n) in the worst case.
        numsLength = len(nums)
        if numsLength < 2:
            return 0
        nums = self.bucketSort(nums)
        l, r = 0, 1
        maxDiff = 0
        while r < numsLength:
            if nums[r] - nums[l] > maxDiff:
                maxDiff = nums[r] - nums[l]
            l += 1
            r += 1
        return maxDiff

    # Time Complexity:
    # Best Case: O(n)
    # Average Case: O(n + k log k)
    # Worst Case: O(n log n)

    # Space Complexity: 
    # O(n)

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