updates

Author: rusuraluca

Data Structures/HashMap/ContainsDuplicateII.py

@@ -0,0 +1,15 @@
+"""
+Problem:
+-----------------------------------------------
+https://leetcode.com/problems/contains-duplicate-ii/
+"""
+
+
+class Solution:
+    def containsNearbyDuplicate(self, nums, k: int) -> bool:
+        seen = {}
+        for i, n in enumerate(nums):
+            if n in seen and i - seen[n] <= k:
+                return True
+            seen[n] = i
+        return False

Data Structures/HashMap/GroupAnagrams.py

@@ -0,0 +1,56 @@
+"""
+Problem:
+-----------------------------------------------
+https://leetcode.com/problems/group-anagrams
+
+
+HashMap Solution:
+-----------------------------------------------
+@description:
+We can group the anagrams in a hashmap
+where we create a key representing their letters sorted.
+
+Actually, designing a key is to build a mapping relationship by yourself
+between the original information and the actual key used by hash map.
+When you design a key, you need to guarantee that:
+1. All values belong to the same group will be mapped in the same group.
+2. Values which needed to be separated into different groups will not be mapped into the same group.
+
+This process is similar to design a hash function, but here is an essential difference.
+A hash function satisfies the first rule but might not satisfy the second one.
+But your mapping function should satisfy both of them.
+
+In the example above, our mapping strategy can be:
+sort the string and use the sorted string as the key.
+That is to say, both "eat" and "ate" will be mapped to "aet".
+
+Input: strs = ["eat","tea","tan","ate","nat","bat"]
+Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
+
+["eat","tea","tan","ate","nat","bat"]
+
+key       value
+["aet"] - ["eat","tea", "ate"]
+["ant"] - ["nat","tan"]
+["abt"] - ["bat"]
+
+@complexity:
+Time:   O(n*klogk), where n is the number of anagrams and k is the max_len_anagram out of them
+Space:  O(n), for the hashmap of all strings with no repetition + the sorted versions of each anagram as keys
+"""
+class Solution:
+    def groupAnagrams(self, strs):
+        if len(strs) == 0:
+            return [[""]]
+
+        hash_map = {}
+
+        for s in strs:
+            key = ''.join(sorted(s))
+            if key in hash_map:
+                hash_map.get(key).append(s)
+            else:
+                hash_map[key] = [s]
+
+        return hash_map.values()
+