<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 n;
    void bucketsort(vector<int> & nums,long long int dec){
        vector<int> count(10);
        for(auto &i: nums){
            int t=(i/dec)%10;
            // cout<<t<<' ';
            count[t]++;
        }
        for(int i=1;i<10;i++){
            count[i]+=count[i-1];
        }
        vector<int> temp=nums;
        for(int i=n-1;i>=0;i--){
            int t=(temp[i]/(dec))%10;
            nums[--count[t]]=temp[i];
        }
        return;
    }
    int maximumGap(vector<int>& nums) {
        n=nums.size();
        if(n==1) return 0;
        for(long long int dec=1;dec<=1e9;dec*=10){
            bucketsort(nums,dec);
        }
        int ans=0;
        for(int i=1;i<n;i++){
            ans=max(ans,nums[i]-nums[i-1]);
        }
        return ans;
    }
};
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