<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) {

        if (nums.size() == 0 || nums.size() == 1) return 0;
        
        //int i = 0;
        int j = nums.size();
        int firstIdx = -1;
        int lastIdx = -1;
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i+1; j < nums.size(); j++)
            {
                if (nums[i] > nums[j]) 
                {
                    firstIdx = i;
                    break;
                }
            }
            if (firstIdx >= 0) break;
        }

        for (int i = nums.size()-1; i >= 0 && firstIdx >= 0; i--)
        {
            for (int j = i-1; j >= 0; j--)
            {
                if (nums[i] < nums[j]) 
                {
                    lastIdx = i;
                    break;
                }
            }
            if (lastIdx >= 0) break;
        }
        if (lastIdx == -1 || firstIdx == -1) return 0;
        return lastIdx - firstIdx + 1;

    }
};
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