<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) {
    // precompute the array, which tells the minimum value from index = i till the end of the array.
        int arr[100000];
        int n = nums.size();
        arr[n-1] = nums[n-1];
        for(int j = n-2; j>=0; j--){
            arr[j] = min(arr[j+1], nums[j]);
        }

        int max_lp = INT_MIN;
        int min_rp = INT_MAX;
        int partition_size =0;
        for(int i=0; i<n; i++){
            // assume ele upto index = i , are included in left partition
            // ele from i+1 appear in right partition, therefore find the min of them, including the i+1 th ele
            max_lp = max(max_lp, nums[i]);
            min_rp = arr[i+1];
            if(max_lp <= min_rp){
                partition_size = i+1;
                break;
            }
        }
        return partition_size;
    }
};
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