<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 radixsort(vector<int>& nums){
        int exp = 1;
        for(int i=0 ; i<9; i++){
            countSort(nums,exp);
            exp*=10;
        }
    }
    void countSort(vector<int>& nums, int exp){

        int count[10] = {0};
        for(int i=0 ; i<nums.size() ; i++){
           count[(nums[i]/exp) % 10]++; 
        }
        for(int i=1 ; i<10 ; i++){
            count[i] += count[i-1];
        }
        vector<int> output(nums.size());
        for(int i=nums.size()-1 ; i>=0 ; i--){
            output[--count[(nums[i]/exp) % 10]] = nums[i];
        }
        for(int i=0 ; i<output.size() ; i++) nums[i] = output[i];
    }
    int maximumGap(vector<int>& nums) {
        radixsort(nums);
        int maxi = 0;
        for(int i=1 ; i<nums.size() ; i++) maxi = max(maxi, nums[i] - nums[i-1]);
        return maxi;
    }
};
107221