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
using pii = pair<int, int>;

class Solution {
public:
    const int inf = 0x3f3f3f3f;
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) return 0;
        int mi = inf, mx = -inf;
        for (int v : nums) {
            mi = min(mi, v);
            mx = max(mx, v);
        }
        int bucketSize = max(1, (mx - mi) / (n - 1));
        int bucketCount = (mx - mi) / bucketSize + 1;
        vector<pii> buckets(bucketCount, {inf, -inf});
        for (int v : nums) {
            int i = (v - mi) / bucketSize;
            buckets[i].first = min(buckets[i].first, v);
            buckets[i].second = max(buckets[i].second, v);
        }
        int ans = 0;
        int prev = inf;
        for (auto [curmin, curmax] : buckets) {
            if (curmin > curmax) continue;
            ans = max(ans, curmin - prev);
            prev = curmax;
        }
        return ans;
    }
};