In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.

Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographicaly in this alien language.

Example 1:

Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.

Example 2:

Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.

Example 3:

Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).

Note:

1. 1 <= words.length <= 100
2. 1 <= words[i].length <= 20
3. order.length == 26
4. All characters in words[i] and order are english lowercase letters.

Solution: Hashtable

Time complexity: O(sum(len(words[i])))

Space complexity: O(26)

// Author: Huahua, 8 ms
class Solution {
public:
  bool isAlienSorted(vector<string>& words, string order) {
    vector<char> m(26);
    for (int i = 0; i < 26; ++i)
      m[order[i] - 'a'] = 'a' + i;
    for (int i = 0; i < words.size(); ++i) {      
      for (int j = 0; j < words[i].length(); ++j)
        words[i][j] = m[words[i][j] - 'a'];      
      if (i > 0 && words[i] < words[i - 1]) return false;      
    }
    return true;        
  }
};

The post 花花酱 LeetCode 953. Verifying an Alien Dictionary appeared first on Huahua's Tech Road.

Problem:

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].

Idea: 

Reduce the problem to 花花酱 769. Max Chunks To Make Sorted by creating a mapping from number to index in the sorted array.

arr = [2, 3, 5, 4, 4]

sorted = [2, 3, 4, 4, 5]

indices = [0, 1, 4, 2, 3]

Solution: Mapping

Time complexity: O(nlogn)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 19 ms
class Solution {
public:
  int maxChunksToSorted(vector<int>& arr) {
    vector<int> v(arr.size());
    std::iota(v.begin(), v.end(), 0);
    std::sort(v.begin(), v.end(), [&arr](const int i1, const int i2){
      return arr[i1] == arr[i2] ? (i1 < i2) : arr[i1] < arr[i2];
    });
    int ans = 0;
    int m = 0;
    for (int i = 0; i < v.size(); ++i) {
      m = max(m, v[i]);
      if (m == i) ++ans;
    }
    return ans;
  }
};

Python3

"""
Author: Huahua
Running time: 95 ms
"""
class Solution:
  def maxChunksToSorted(self, arr):
    v = sorted([(n << 32) | i for i, n in enumerate(arr)])    
    m = 0
    ans = 0
    for i, n in enumerate(v):
      m = max(m, n & 0xffffffff)
      if i == m: ans += 1    
    return ans

The post 花花酱 LeetCode 768. Max Chunks To Make Sorted II appeared first on Huahua's Tech Road.