<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?
bool compare(const pair<int,int>&p1, const pair<int,int>&p2)
{
    if(p1.first != p2.first)
    {
        return p1.first < p2.first;
    }
    return p1.second < p2.second;
}

class Solution {
public:
int firstUnsortedIndex(vector<pair<int,int>>&v, int i, int n)
{
    while(i<n && v[i].second == i)++i;
    return i;
}

int lastUnsortedIndex(vector<pair<int,int>>&v, int i, int n)
{
    while(i>n && v[i].second == i)--i;
    return i;
}
    int findUnsortedSubarray(vector<int>& nums) {
        int n = nums.size();
        vector<pair<int,int>>v(n);
        for(int i=0;i<n;++i)
        {
            v[i].first = nums[i];
            v[i].second = i;
        }
        sort(v.begin(), v.end(), compare);
        int fI = firstUnsortedIndex(v,0,n);
        int lI = lastUnsortedIndex(v,n-1,fI);
        return lI-fI+1;
    }
};
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