<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 minNum=*min_element(nums.begin(),nums.end());
        int maxNum=*max_element(nums.begin(),nums.end());

        int bucketSize=max(1,(maxNum-minNum)/(n-1));

        vector<vector<int>> buckets((maxNum-minNum)/bucketSize+1);

        for(int num:nums){
            int index=(num-minNum)/bucketSize;
            if(buckets[index].empty()){
                buckets[index]={num,num};
            } else{
                buckets[index][0]=min(buckets[index][0],num);
                buckets[index][1]=max(buckets[index][1],num);
            }
        }
        int maxGap=0;
        int previousMax=minNum;
        for(const auto&bucket:buckets){
            if(bucket.empty()) continue;
            maxGap=max(maxGap,bucket[0]-previousMax);
            previousMax=bucket[1];
        }
        return maxGap;
    }
};
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