<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:
    static bool compare(pair<int,int>& a,pair<int,int>& b){
        if(a.first!=b.first)return a.first<b.first;
        return a.second<b.second;
    }
    int maxChunksToSorted(vector<int>& arr) {
        int n=arr.size();
        vector<pair<int,int>> v(n);
        for(int i=0;i<n;i++)v[i]={arr[i],i};
        sort(v.begin(),v.end(),compare);
        map<int,int> mp;
        for(int i=0;i<n;i++){
            mp.insert({v[i].second,i});
        }
        // for(auto x:mp){
        //     cout<<x.first<<" "<<x.second<<endl;
        // }
        vector<int> vis(n,0);
        int maxidx=0,l=0;
        int ans=0;
        for(int i=0;i<n;i++){
            vis[mp[i]]=1;
            maxidx=max(maxidx,mp[i]);
            int ct=0;
            for(int j=l;j<=maxidx;j++){
                if(vis[j]==1)ct++;
            }
            if(ct==maxidx-l+1){
                ans++;
                l=maxidx+1;
            }
        }
        return ans;
    }
};
19200