<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:
    void countsort(vector<int> &nums,int place){
        vector<int>temp(10,0);
        vector<int>res(nums.size());
        for(int i=0;i<nums.size();i++){
            temp[(nums[i]/place)%10]++;
        }
        for(int i=1;i<temp.size();i++){
            temp[i]+=temp[i-1];

        }
        for(int i=res.size()-1;i>=0;i--){
            res[temp[(nums[i]/place)%10]-1]=nums[i];
            temp[(nums[i]/place)%10]--;
        }
        for(int i=0;i<res.size();i++){
            nums[i]=res[i];
        }

    }
    void radixsort(vector<int> &nums){
        int max=*max_element(nums.begin(),nums.end());
        for(int place=1;(max/place)!=0;place*=10){
            countsort(nums,place);

        }
    }
    int maximumGap(vector<int>& nums){
        radixsort(nums);
        int diff=0;
        for(int i=1;i<nums.size();i++){
            diff=max(diff,(nums[i]-nums[i-1]));
        }
        return diff;
        
    }
};
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