<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>& arr) {
    const int n = arr.size();
    int ans = 0;
    vector<int> maxL(n);  
    vector<int> minR(n);  

    for (int i = 0; i < n; ++i)
      maxL[i] = i == 0 ? arr[i] : max(arr[i], maxL[i - 1]);

    for (int i = n - 1; i >= 0; --i)
      minR[i] = i == n - 1 ? arr[i] : min(arr[i], minR[i + 1]);

    for (int i = 0; i + 1 < n; ++i)
      if (maxL[i] <= minR[i + 1])
        ++ans;

    return ans + 1;
  }
};
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