<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) {
        set<int> setOfElements;
        if(nums.size() < 2) return 0;
        for(int i=0;i<nums.size();i++) {
            setOfElements.insert(nums[i]);
        }

        int flag = 0;
        int element;
        int maxGap = 0;
        for(auto &it:setOfElements) {
            if(flag == 0) {
                element = it;
                flag = 1;
            } else {
                int gap = it - element;
                if(gap > maxGap) {
                    maxGap = gap;
                }
                element = it;
            }
        }
        return maxGap;
    }
};
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