From 25fe22ae23f9d08b64f15540da2a1690f1727ab4 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Thu, 21 Apr 2022 14:16:42 +0530
Subject: [PATCH] add group-anagrams

---
 0049_group-anagrams/README.md           | 27 +++++++++++++++++++++++++
 0049_group-anagrams/python3/naive.py    | 13 ++++++++++++
 0049_group-anagrams/python3/solution.py | 24 ++++++++++++++++++++++
 3 files changed, 64 insertions(+)
 create mode 100644 0049_group-anagrams/README.md
 create mode 100644 0049_group-anagrams/python3/naive.py
 create mode 100644 0049_group-anagrams/python3/solution.py

diff --git a/0049_group-anagrams/README.md b/0049_group-anagrams/README.md
new file mode 100644
index 0000000..75088df
--- /dev/null
+++ b/0049_group-anagrams/README.md
@@ -0,0 +1,27 @@
+Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.
+
+An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+**Example 1:**
+
+    Input: strs = ["eat","tea","tan","ate","nat","bat"]
+    Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
+    
+
+**Example 2:**
+
+    Input: strs = [""]
+    Output: [[""]]
+    
+
+**Example 3:**
+
+    Input: strs = ["a"]
+    Output: [["a"]]
+    
+
+**Constraints:**
+
+*   `1 <= strs.length <= 104`
+*   `0 <= strs[i].length <= 100`
+*   `strs[i]` consists of lowercase English letters.
\ No newline at end of file
diff --git a/0049_group-anagrams/python3/naive.py b/0049_group-anagrams/python3/naive.py
new file mode 100644
index 0000000..c05bf38
--- /dev/null
+++ b/0049_group-anagrams/python3/naive.py
@@ -0,0 +1,13 @@
+# Time: O(N * MlogM) where N is len(strs) and M is num of chars in a string
+# Space: O(N)
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        groups = {}
+        
+        for s in strs:
+            s_sorted = ''.join(sorted(s))
+            
+            groups[s_sorted] = groups.get(s_sorted, []) + [s]
+        
+        return list(groups.values())
diff --git a/0049_group-anagrams/python3/solution.py b/0049_group-anagrams/python3/solution.py
new file mode 100644
index 0000000..459b7d4
--- /dev/null
+++ b/0049_group-anagrams/python3/solution.py
@@ -0,0 +1,24 @@
+# Time: O(N · M)
+# Space: O(N)
+
+from collections import Counter, defaultdict
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        # If key doesn't exist, creates empty list using `list`
+        # factory/ctor function
+        groups = defaultdict(list)
+        
+        for s in strs:
+            # Keep counter for num of times each letter appeared
+            counts = [0] * 26
+            
+            for c in s:
+                i = ord(c) - ord('a')
+                counts[i] += 1
+            
+            # Lists aren't hashable, so we need to convert the counts
+            # to tuple so that we can key them
+            groups[tuple(counts)].append(s)
+        
+        return list(groups.values())