<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:
    vector<int> countsort(vector<int>& nums) {
        int n = nums.size();
        map<int, int> m;
        for (int i = 0; i < n; i++)
            m[nums[i]]++;

        nums.clear();
        for (auto it = m.begin(); it != m.end(); it++) {
            int cnt = it->second;
            while (cnt--) {
                nums.push_back(it->first);
            }
        }
        return nums;
    }
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2)
            return 0;
        vector<int> sortednums = countsort(nums);

        int maxgap = INT_MIN;
        for (int i = 1; i < n; i++) {
            int currgap = sortednums[i] - sortednums[i - 1];
            maxgap = max(maxgap, currgap);
        }

        return maxgap;
    }
};
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