<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 maxi=*max_element(nums.begin(),nums.end());
        int mini=*min_element(nums.begin(),nums.end());

        int n=nums.size();
        if(n==1)return 0;
        int sz=(maxi-mini)/(n-1);
        if(sz*(n-1)!=maxi-mini)sz++;

        //sz=max(sz,1); ////////all elements are same
        if(sz==0)return 0;

        vector<int>b[n];

        for(int i=0;i<n;i++){
            int idx=(nums[i]-mini)/sz;
            b[idx].push_back(nums[i]);
        }
        
       // cout<<sz<<endl;
        int ans=sz;
        int prevmaxi=-1;
        for(int i=0;i<n;i++){
            int curmaxi=-1,curmini=2e9;
            for(auto it:b[i]){
                curmaxi=max(curmaxi,it);
                curmini=min(curmini,it);
            }
            if(prevmaxi!=-1&&curmini!=2e9){
                ans=max(ans,curmini-prevmaxi);
            }
            prevmaxi=max(prevmaxi,curmaxi);
        }

        return ans;
    }
};
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