[LeetCode] 768. Max Chunks To Make Sorted II 可排序的最大块数 II

This question is the same as "Max Chunks to Make Sorted" 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.

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].

769. Max Chunks To Make Sorted的拓展，这一题可以有重复的数字。

解法：能形成块儿的数字之和跟排序后的数组的相同长度的子数组的数字之和是相同的。

Algorithm: Iterate through the array, each time all elements to the left are smaller (or equal) to all elements to the right, there is a new chunck.
Use two arrays to store the left max and right min to achieve O(n) time complexity. Space complexity is O(n) too.
This algorithm can be used to solve ver1 too.

Java:

class Solution {
    public int maxChunksToSorted(int[] arr) {
        int n = arr.length;
        int[] maxOfLeft = new int[n];
        int[] minOfRight = new int[n];
 
        maxOfLeft[0] = arr[0];
        for (int i = 1; i < n; i++) {
            maxOfLeft[i] = Math.max(maxOfLeft[i-1], arr[i]);
        }
 
        minOfRight[n - 1] = arr[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            minOfRight[i] = Math.min(minOfRight[i + 1], arr[i]);
        }
 
        int res = 0;
        for (int i = 0; i < n - 1; i++) {
            if (maxOfLeft[i] <= minOfRight[i + 1]) res++;
        }
 
        return res + 1;
    }
}　　

Python:

def maxChunksToSorted(self, arr):
        res, c1, c2 = 0, collections.Counter(), collections.Counter()
        for a, b in zip(arr, sorted(arr)):
            c1[a] += 1
            c2[b] += 1
            res += c1 == c2
        return res

C++:

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int res = 0, sum1 = 0, sum2 = 0;
        vector<int> expect = arr;
        sort(expect.begin(), expect.end());
        for (int i = 0; i < arr.size(); ++i) {
            sum1 += arr[i];
            sum2 += expect[i];
            if (sum1 == sum2) ++res;
        }
        return res;
    }
};

C++:

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int res = 1, n = arr.size();
        vector<int> f = arr, b = arr;   
        for (int i = 1; i < n; ++i) f[i] = max(arr[i], f[i - 1]);   
        for (int i = n - 2; i >= 0; --i) b[i] = min(arr[i], b[i + 1]);
        for (int i = 0; i < n - 1; ++i) {
            if (f[i] <= b[i + 1]) ++res;
        }
        return res;
    }
};

C++:

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int res = 1, n = arr.size(), curMax = INT_MIN;
        vector<int> b = arr;    
        for (int i = n - 2; i >= 0; --i) b[i] = min(arr[i], b[i + 1]);
        for (int i = 0; i < n - 1; ++i) {
            curMax = max(curMax, arr[i]);
            if (curMax <= b[i + 1]) ++res;
        }
        return res;
    }
};

C++:

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        stack<int> st;
        for (int i = 0; i < arr.size(); ++i) {
            if (st.empty() || st.top() <= arr[i]) {
                st.push(arr[i]);
            } else {
                int curMax = st.top(); st.pop();
                while (!st.empty() && st.top() > arr[i]) st.pop();
                st.push(curMax);
            }
        }
        return st.size();
    }
};