<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
        
        # Find the minimum and maximum of the array
        ma, mi = max(nums), min(nums)
        
        # If all numbers are the same, return 0 as there is no gap
        if ma == mi:
            return 0
        
        # Calculate the size of each bucket
        bucket_size = max(1, (ma - mi) // (n - 1))  # Use max to avoid zero bucket size
        bucket_count = (ma - mi) // bucket_size + 1  # The number of buckets
        
        # Initialize buckets as pairs [min, max]
        buckets = [[math.inf, -math.inf] for _ in range(bucket_count)]
        
        # Place each number in its corresponding bucket
        for x in nums:
            # Calculate bucket index
            i = (x - mi) // bucket_size
            buckets[i][0] = min(buckets[i][0], x)  # Update bucket's min
            buckets[i][1] = max(buckets[i][1], x)  # Update bucket's max
        
        # Initialize the max_diff to 0
        max_diff = 0
        last = mi  # Start with the minimum number
        
        # Iterate over the buckets to find the maximum gap between consecutive buckets
        for i in range(bucket_count):
            # Skip empty buckets
            if buckets[i][0] == math.inf and buckets[i][1] == -math.inf:
                continue
            
            # The gap is the difference between the minimum of the current bucket
            # and the maximum of the previous bucket
            max_diff = max(max_diff, buckets[i][0] - last)
            
            # Update last to the current bucket's maximum
            last = buckets[i][1]
        
        return max_diff
35203