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?
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length())
return false;
char[] str1 = s.toCharArray();
char[] str2 = t.toCharArray();
Arrays.sort(str1);
Arrays.sort(str2);
return Arrays.equals(str1, str2);
}
}class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length())
return false;
int[] chars = new int[26];
for (int i = 0; i < s.length(); i++)
chars[s.charAt(i) - 'a']++;
for (int i = 0; i < t.length(); i++) {
chars[t.charAt(i) - 'a']--;
if (chars[t.charAt(i) - 'a'] < 0) {
return false;
}
}
return true;
}
}What if the inputs contain unicode characters? How would you adapt your solution to such case?
Answer : Use a hash table instead of a fixed size counter. Imagine allocating a large size array to fit the entire range of unicode characters, which could go up to more than 1 million. A hash table is a more generic solution and could adapt to any range of characters.