题目描述：

LeetCode 768. Max Chunks To Make Sorted (ver. 2)

This question is the same as "Max Chunks to Make Sorted (ver. 1)" except the integers of the given array are not necessarily distinct, the input array could be up to length 2000, and the elements could be up to 10**8.

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Given an array arr of integers (not necessarily distinct), we split the array into some number of "chunks" (partitions), and individually sort each chunk.  After concatenating them, the result equals the sorted array.

What is the most number of chunks we could have made?

Example 1:

Input: arr = [5,4,3,2,1]
Output: 1
Explanation:
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't sorted.

Example 2:

Input: arr = [2,1,3,4,4]
Output: 4
Explanation:
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.

Note:

• arr will have length in range [1, 2000].
• arr[i] will be an integer in range [0, 10**8].

题目大意：

该题目是Max Chunks to Make Sorted (ver. 1)的变体，区别是元素不一定唯一，数组长度2000以内，元素大小10**8。

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将数组arr（元素可以重复）拆分成若干部分，使得每一部分分别排序后，重新组合在一起的数组非递减有序。

求最多可以分成多少个部分。

解题思路：

贪心算法（Greedy Algorithm）

记数组arr长度为N，下标x从N - 1到0逐一递减：

    若min(arr[x .. N - 1]) >= max(arr[0 .. x - 1])，则令结果+1

Python代码：

class Solution(object):
    def maxChunksToSorted(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        N = len(arr)
        ans = 1
        for x in range(N - 1, 0, -1):
            if min(arr[x:]) >= max(arr[:x]):
                ans += 1
        return ans

 

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