<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 maximumGap(vector<int>& nums) {
        if (nums.size() < 2) return 0;  

        int n = nums.size();

        map<int, int> freq;
        for (auto it : nums) {
            freq[it]++;
        }

        nums.clear();
        for(auto it : freq) {
            while(it.second--) {
                nums.push_back(it.first);
            }
        }

        int maxGap = INT_MIN;
        for (int i = 1; i < nums.size(); i++) {
            maxGap = max(maxGap, nums[i] - nums[i - 1]);
        }

        return maxGap;
    }
};

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