<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 {
public:
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) return 0;

        int minElem = *min_element(nums.begin(), nums.end());
        int maxElem = *max_element(nums.begin(), nums.end());

        if (minElem == maxElem) return 0;

        int bucketSize = max(1, (maxElem - minElem) / (n - 1));
        int bucketCount = (maxElem - minElem) / bucketSize + 1;

        vector<int> bucketMin(bucketCount, INT_MAX);
        vector<int> bucketMax(bucketCount, INT_MIN);

        for (int num : nums) {
            int idx = (num - minElem) / bucketSize;
            bucketMin[idx] = min(bucketMin[idx], num);
            bucketMax[idx] = max(bucketMax[idx], num);
        }

        int maxGap = 0, prev = minElem;
        for (int i = 0; i < bucketCount; ++i) {
            if (bucketMin[i] == INT_MAX) continue;
            maxGap = max(maxGap, bucketMin[i] - prev);
            prev = bucketMax[i];
        }

        return maxGap;
    }
};
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