<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) {
        map<int,int> mp, mm;
        vector<int> tmp = arr;
        sort(tmp.begin(),tmp.end());
        int n = arr.size(), l = 1e9, r = -1e9;

        for(int i = 0; i < n;i++){
            if(arr[i]==tmp[i]){
                if(mp.find(arr[i]) != mp.end()){
                    continue;
                }
            }
            mp[arr[i]] = i;
        }
        int ans = 0;
        for(int i = 0; i < n;i++){
            if(arr[i]==tmp[i]){
                if((l==1e9 && r == -1e9)){
                    ans++;
                    l = 1e9, r = -1e9;
                }
            }
            else if(r == i){
                ans++;
                l = 1e9, r = -1e9;
            }
            else{
                int a = lower_bound(tmp.begin(), tmp.end(), arr[i]) - tmp.begin() + mm[arr[i]];
                int b = mp[tmp[i]];
                l = min(l, min(a, min(b, i)));
                r = max(r, max(a, max(b, i)));
            }
            mm[arr[i]]++;
        }
        if(l != 1e9 || r != -1e9){
            return ans+1;
        }
        return min(n,ans);
    }
};
20000