Maximum Gap
<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>
Array; Bucket Sort; Radix Sort; Sorting
func maximumGap(nums []int) int {
	n := len(nums)
	if n < 2 {
		return 0
	}
	inf := 0x3f3f3f3f
	mi, mx := inf, -inf
	for _, v := range nums {
		mi = min(mi, v)
		mx = max(mx, v)
	}
	bucketSize := max(1, (mx-mi)/(n-1))
	bucketCount := (mx-mi)/bucketSize + 1
	buckets := make([][]int, bucketCount)
	for i := range buckets {
		buckets[i] = []int{inf, -inf}
	}
	for _, v := range nums {
		i := (v - mi) / bucketSize
		buckets[i][0] = min(buckets[i][0], v)
		buckets[i][1] = max(buckets[i][1], v)
	}
	ans := 0
	prev := inf
	for _, bucket := range buckets {
		if bucket[0] > bucket[1] {
			continue
		}
		ans = max(ans, bucket[0]-prev)
		prev = bucket[1]
	}
	return ans
}