Leetcode - Strings Simple

##49. Group Anagrams[M]

https://leetcode.com/problems/group-anagrams/

Description

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

All inputs will be in lowercase.
The order of your output does not matter.

Solution

https://leetcode.com/problems/group-anagrams/solution/

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        strs.sort()
        hash = {}
        for s in strs:
            key = self.hash_key(s)
            try:
                hash[key].append(s)
            except KeyError:
                hash[key] = [s]
        return hash.values()
    def hash_key(self, s):
        # hash string with 26 length array
        table = [0] * 26
        for ch in s:
            index = ord(ch) - ord('a')
            table[index] += 1
        return str(table)

706. Design HashMap[E]

https://leetcode.com/problems/design-hashmap/

Description

Design a HashMap without using any built-in hash table libraries.

To be specific, your design should include these functions:

put(key, value) : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
get(key): Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
remove(key) : Remove the mapping for the value key if this map contains the mapping for the key.

Example:

MyHashMap hashMap = new MyHashMap();
hashMap.put(1, 1);          
hashMap.put(2, 2);         
hashMap.get(1);            // returns 1
hashMap.get(3);            // returns -1 (not found)
hashMap.put(2, 1);          // update the existing value
hashMap.get(2);            // returns 1 
hashMap.remove(2);          // remove the mapping for 2
hashMap.get(2);            // returns -1 (not found)

Note: