<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 pos(int x, int fact) {
        return ((int)(x / pow(10, fact)) % 10);
    }

    void countSort(vector<int>& nums, int fact) {
        int N = (int)nums.size();
        vector<int> count(10, 0);
        for (int i : nums) 
            count[pos(i, fact)]++;
        for (int i=1; i<10; i++)
            count[i] += count[i-1];
        
        vector<int> sort(N, 0);
        for (int i=N-1; i>=0; i--) 
            sort[(count[pos(nums[i], fact)]--)-1] = nums[i];
        nums = sort;
    }

    int maximumGap(vector<int>& nums) {
        int N = (int)nums.size();
        int MN = *min_element(nums.begin(), nums.end());
        for (int& i : nums) i += MN;
        int MX = *max_element(nums.begin(), nums.end());

        for (int cnt=0; cnt<log10(MX)+1; cnt++) 
            countSort(nums, cnt);

        int ans = 0;
        for (int i=1; i<N; i++)
            ans = max(ans, nums[i] - nums[i-1]);
        
        return ans;
    }
};
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