<p>You are given an integer array <code>arr</code>.</p>

<p>We split <code>arr</code> into some number of <strong>chunks</strong> (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.</p>

<p>Return <em>the largest number of chunks we can make to sort the array</em>.</p>

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

<pre>
<strong>Input:</strong> arr = [5,4,3,2,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
Splitting into two or more chunks will not return the required result.
For example, splitting into [5, 4], [3, 2, 1] will result in [4, 5, 1, 2, 3], which isn&#39;t sorted.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,1,3,4,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong>
We can split into two chunks, such as [2, 1], [3, 4, 4].
However, splitting into [2, 1], [3], [4], [4] is the highest number of chunks possible.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 2000</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution {
public:

    int maxChunksToSorted(vector<int>& a) 
    {

        int n=a.size();
        
        vector<int>preMax(n,0),sufMin(n,INT_MAX);
        int ans=1;
        preMax[0]=a[0];
        sufMin[n-1]=a[n-1];

        for(int i=1;i<n;i++)
        {
            preMax[i]=max(preMax[i-1],a[i]);
        }
        for(int i=n-2;i>=0;i--)
        {
            sufMin[i]=min(sufMin[i+1],a[i]);
        }

        for(int i=0;i<n-1;i++)
        {
            if(preMax[i]<=sufMin[i+1])
            {
                ans++;
            }
        }

        return ans;
    }
};
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