<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 i=0;
        int j=n-1;
        while(i<n){
            if(i<n-1 && nums[i]>nums[i+1]){
                // i decided
                break;
            }
            i++;
        }
        while(j>=0){
            if(j>0 && nums[j]<nums[j-1]){
                //j decided
                break;
            }
            j--;
        }
        if (i > j) return 0;
        int mini=INT_MAX;
        int maxi=INT_MIN;
        for(int k=i;k<=j;k++){
            mini = min(mini,nums[k]);
            maxi = max(maxi,nums[k]);
        }
        int s=i;
        int e=j;
        for(int k=0;k<i;k++){
            if(nums[k]>mini){
                s=k;
                break;
            }
        }
        for(int k=n-1;k>j;k--){
            if(nums[k]<maxi){
                e=k;
                break;
            }
        }
        return e-s+1;
        
    }
};
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