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
class Solution {
    public int maximumGap(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }
        int inf = 0x3f3f3f3f;
        int mi = inf, mx = -inf;
        for (int v : nums) {
            mi = Math.min(mi, v);
            mx = Math.max(mx, v);
        }
        int bucketSize = Math.max(1, (mx - mi) / (n - 1));
        int bucketCount = (mx - mi) / bucketSize + 1;
        int[][] buckets = new int[bucketCount][2];
        for (var bucket : buckets) {
            bucket[0] = inf;
            bucket[1] = -inf;
        }
        for (int v : nums) {
            int i = (v - mi) / bucketSize;
            buckets[i][0] = Math.min(buckets[i][0], v);
            buckets[i][1] = Math.max(buckets[i][1], v);
        }
        int prev = inf;
        int ans = 0;
        for (var bucket : buckets) {
            if (bucket[0] > bucket[1]) {
                continue;
            }
            ans = Math.max(ans, bucket[0] - prev);
            prev = bucket[1];
        }
        return ans;
    }
}