<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:
    void bs(vector<int>& cache, int num) {
        // find the first one that it could go to
        int left = 0, right = cache.size() - 1, ans = 0;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (cache[mid] <= num) {
                left = mid + 1;
            } else {
                ans = mid;
                right = mid - 1;
            }
        }
        int maximum = cache.back();
        int n = cache.size();
        for (int i = n - 1; i >= ans; --i) {
            cache.pop_back();
        }
        cache.push_back(maximum);
    }
    int maxChunksToSorted(vector<int>& arr) {
        // everything in this chunk should be less than or equal to everything in the chunk after it and so on
        // greedy approach

        // keep track of the largest in each chunk
        // binary search to find the chunk you need

        vector<int> largest;
        largest.push_back(arr[0]);
        int prev = arr[0];
        for (int i = 1; i < arr.size(); ++i) {
            int num = arr[i];
            if (num < prev) {
                // binary search for first instance of prev
                bs(largest, num);
            } else {
                largest.push_back(num);
            }
            prev = largest.back();
        }

        return largest.size();
    }
};
16000