add valid-anagram

0242_valid-anagram/README.md

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+Given two strings `s` and `t`, return `true` _if_ `t` _is an anagram of_ `s`_, and_ `false` _otherwise_.
+
+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: s = "anagram", t = "nagaram"
+    Output: true
+    
+
+**Example 2:**
+
+    Input: s = "rat", t = "car"
+    Output: false
+    
+
+**Constraints:**
+
+*   `1 <= s.length, t.length <= 5 * 104`
+*   `s` and `t` consist of lowercase English letters.
+
+**Follow up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

0242_valid-anagram/python3/nlogn.py

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+# Time: O(NlogN)
+# Space: O(1)
+
+from collections import Counter
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        slen, tlen = len(s), len(t)
+        
+        if slen != tlen:
+            return False
+        
+        return sorted(s) == sorted(t)

0242_valid-anagram/python3/oneliner.py

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+# Time: O(N)
+# Space: O(N)
+
+from collections import Counter
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        return Counter(s) == Counter(t)

0242_valid-anagram/python3/solution.py

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+# Time: O(N) where N is num of chars in string
+# Space: O(N) where N is num of chars in string
+
+from collections import Counter
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        slen, tlen = len(s), len(t)
+        
+        if slen != tlen:
+            return False
+        
+        scounts, tcounts = {}, {}
+        
+        for i in range(slen):
+            sc, tc = s[i], t[i]
+            
+            scounts[sc] = scounts.get(sc, 0) + 1
+            tcounts[tc] = tcounts.get(tc, 0) + 1
+            
+        return scounts == tcounts