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Author: BaffinLee

201-300/242. Valid Anagram.md

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+# 242. Valid Anagram
+
+- Difficulty: Easy.
+- Related Topics: Hash Table, Sort.
+- Similar Questions: Group Anagrams, Palindrome Permutation, Find All Anagrams in a String.
+
+## Problem
+
+Given two strings **s** and **t **, write a function to determine if **t** is an anagram of **s**.
+
+**Example 1:**
+
+```
+Input: s = "anagram", t = "nagaram"
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: s = "rat", t = "car"
+Output: false
+```
+
+**Note:**
+You may assume the string contains only lowercase alphabets.
+
+**Follow up:**
+What if the inputs contain unicode characters? How would you adapt your solution to such case?
+
+
+## Solution 1
+
+```javascript
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+  var lenA = s.length;
+  var lenB = t.length;
+  var map = {};
+  
+  if (lenA !== lenB) return false;
+  
+  for (var i = 0; i < lenA; i++) {
+    if (!map[s[i]]) map[s[i]] = 0;
+    map[s[i]]++;
+  }
+  
+  for (var j = 0; j < lenB; j++) {
+    if (!map[t[j]]) return false;
+    map[t[j]]--;
+  }
+  
+  return true;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).
+
+## Solution 2
+
+```javascript
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+  var lenA = s.length;
+  var lenB = t.length;
+  var map = Array(26);
+  var index = 0;
+  var base = 'a'.charCodeAt(0);
+  
+  if (lenA !== lenB) return false;
+  
+  for (var i = 0; i < lenA; i++) {
+    index = s[i].charCodeAt(0) - base;
+    if (!map[index]) map[index] = 0;
+    map[index]++;
+  }
+  
+  for (var j = 0; j < lenB; j++) {
+    index = t[j].charCodeAt(0) - base;
+    if (!map[index]) return false;
+    map[index]--;
+  }
+  
+  return true;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).