class Solution(object):
    def maxChunksToSorted(self, arr):
        n= len(arr)

        count=0
        currentmax= -2**63
        for i in range(0,n):
            currentmax=max(currentmax, arr[i])
            if (currentmax==i):
                count+=1
        return count
        
        
    
class Solution {
    public int maxChunksToSorted(int[] arr) {

        int max=0, count=0;
        for(int i=0; i<arr.length; i++){
            max = Math.max(arr[i],max);
            if(i==max) count++;
        }return count;
    }
}
Max Chunks To Make Sorted
You are given an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1].

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

Return the largest number of chunks we can make to sort the array.

&nbsp;
Example 1:

Input: arr = [4,3,2,1,0]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.


Example 2:

Input: arr = [1,0,2,3,4]
Output: 4
Explanation:
We can split into two chunks, such as [1, 0], [2, 3, 4].
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.


&nbsp;
Constraints:


	n == arr.length
	1 &lt;= n &lt;= 10
	0 &lt;= arr[i] &lt; n
	All the elements of arr are unique.


var maxChunksToSorted = function(arr) {
    let count = 0, cumSum = 0;
    arr.forEach((el, index) => {
        cumSum += el-index;
        if (cumSum === 0) count++;
    });
    return count;
};
class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        // using chaining technique
        int maxi=INT_MIN;
        int ans =0;

        for(int i=0;i<size(arr);i++)

        {
            maxi = max(maxi,arr[i]);

            if(maxi==i)
            {
                //.. found partition
                ans++;

            }
        }
        return ans;
    }
};