<p>Given an integer array <code>nums</code>, partition it into two (contiguous) subarrays <code>left</code> and <code>right</code> so that:</p>

<ul>
	<li>Every element in <code>left</code> is less than or equal to every element in <code>right</code>.</li>
	<li><code>left</code> and <code>right</code> are non-empty.</li>
	<li><code>left</code> has the smallest possible size.</li>
</ul>

<p>Return <em>the length of </em><code>left</code><em> after such a partitioning</em>.</p>

<p>Test cases are generated such that partitioning exists.</p>

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

<pre>
<strong>Input:</strong> nums = [5,0,3,8,6]
<strong>Output:</strong> 3
<strong>Explanation:</strong> left = [5,0,3], right = [8,6]
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,0,6,12]
<strong>Output:</strong> 4
<strong>Explanation:</strong> left = [1,1,1,0], right = [6,12]
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li>There is at least one valid answer for the given input.</li>
</ul>

class Solution {
public:
    int partitionDisjoint(vector<int>& nums) {
        // 5 5 5 8 8
        // 0 0 3 6 6
        int n=nums.size();
        vector<int> pref(n,0);//largest possible element in left
        vector<int> suff(n,0);//smallest possible element in right
        pref[0]=nums[0];
        for(int i=1;i<n;i++)
            pref[i]=max(pref[i-1],nums[i]);
        suff[n-1]=nums[n-1];
        for(int i=n-2;i>=0;i--)
            suff[i]=min(nums[i],suff[i+1]);
        int ind=0;
        for(int i=0;i<n-1;i++)
        {
            if(pref[i]<=suff[i+1])
            {
                ind=i;
                break;
            }
        }
        return ind+1;
    }
};
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