LeetCode 769. Max Chunks To Make Sorted

原题链接在这里：https://leetcode.com/problems/max-chunks-to-make-sorted/

题目：

Given an array arr that is a permutation of [0, 1, ..., arr.length - 1], 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 = [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.

Note:

• arr will have length in range [1, 10].
• arr[i] will be a permutation of [0, 1, ..., arr.length - 1].

 

题解：

From examples, we could tell if index moves to a point that all the value on its left including itself could cover all the indices, then it could be split into a chunk.

[2, 0, 1]. when i = 0, all the values [2], index i = 0 is not covered by 0, then it is not a chunk.

i = 1. values [2, 0], i = 1 is not covered by 1, then it is not a chunk.

i = 2. values [2, 0, 1] all indices get covered. It could be split into a chunk.

Then how to tell if all the indices are coverred. One way is to maintain the max, if max == i, then all indices are coverred. If max != i, then max must be larger than i. That means there is one value scanned larger, then there must be one index position not scanned yet.

Time Complexity: O(n). n = arr.length. 

Space: O(1).

AC Java:

class Solution {
    public int maxChunksToSorted(int[] arr) {
        int res = 0;
        if(arr == null || arr.length == 0){
            return res;
        }
        
        int max = 0;
        for(int i = 0; i<arr.length; i++){
            max = Math.max(max, arr[i]);
            if(max == i){
                res++;
            }
        }
        
        return res;
    }
}

跟上Max Chunks To Make Sorted II.