<p>Given an integer array <code>nums</code>, you need to find one <b>continuous subarray</b> such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.</p>

<p>Return <em>the shortest such subarray and output its length</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,6,4,8,10,9,15]
<strong>Output:</strong> 5
<strong>Explanation:</strong> You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time complexity?
class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
       
         vector<int>V;

         for(int i = 0;i<nums.size();i++)
         {
            V.push_back(nums[i]);
         
         }

         sort(V.begin(),V.end());

         int first = -1;
         int end = -1;

         for(int i = 0 ; i<nums.size();i++)
         {
            if(nums[i] != V[i])
            {
                if(first == -1)
                {
                    first = i;
                    end = i;
                }
                else
                {
                    end = i;
                }
            }
         }
         if(first == -1)
         {
            return 0;
         }
         return end - first + 1;

    }
};

32700