<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) {
        int n=nums.size();
        int minn=INT_MAX;
        int maxx=INT_MIN;

        if(n==1) return 0;

        for(int i=0;i<n;i++)
        {
            if(i==0)
            {
                if(nums[i]>nums[i+1])
                {
                    minn=min(minn,nums[i]);
                    maxx=max(maxx,nums[i]);
                }
            }
            else if(i==n-1)
            {
                if(nums[i]<nums[i-1])
                {
                    minn=min(minn,nums[i]);
                    maxx=max(maxx,nums[i]);
                }
            }
            else
            {
                if(nums[i]>nums[i+1] || nums[i]<nums[i-1])
                {
                    minn=min(minn,nums[i]);
                    maxx=max(maxx,nums[i]);
                }
            }
        }
        if(minn==INT_MAX || maxx==INT_MIN)
            {
                return 0;
            }

        int i,j;
        for(i=0   ; i<n  && nums[i] <= minn;i++);
        for(j=n-1 ; j>=0 && nums[j] >= maxx ;j--);
        return j-i+1;
    }
};
29200