Given two strings s and t, write a function to determine if t is an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return 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?

Approach #1 (Sorting) [Accepted]

Algorithm

An anagram is produced by rearranging the letters of $s$ into $t$. Therefore, if $t$ is an anagram of $s$, sorting both strings will result in two identical strings. Furthermore, if $s$ and $t$ have different lengths, $t$ must not be an anagram of $s$ and we can return early.

public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) {        //若長度不同則直接返回否
        return false;
    }
    char[] str1 = s.toCharArray();         //String.toCharArray() 方法將字元串轉換為字元陣列
    char[] str2 = t.toCharArray();
    Arrays.sort(str1);                     //Arrays.sort() 需加包import java.util.*;或import java.util.Arrays
 

    Arrays.sort(str2);
    return Arrays.equals(str1, str2);      //Returns true if the two specified arrays of chars are equal to one another
}

Complexity analysis

• Time complexity : $O(n\log n)$. Assume that $n$ is the length of $s$, sorting costs $O(n\log n)$ and comparing two strings costs $O(n)$. Sorting time dominates and the overall time complexity is $O(n\log n)$.

• Space complexity : $O(1)$. Space depends on the sorting implementation which, usually, costs $O(1)$ auxiliary space if heapsort is used. Note that in Java,toCharArray() makes a copy of the string so it costs $O(n)$ extra space, but we ignore this for complexity analysis because:

  • It is a language dependent detail.
  • It depends on how the function is designed. For example, the function parameter types can be changed to char[].

pproach #2 (Hash Table) [Accepted]

Algorithm

To examine if $t$ is a rearrangement of $s$, we can count occurrences of each letter in the two strings and compare them. Since both $s$ and $t$ contain only letters from $a-z$, a simple counter table of size 26 is suffice.

Do we need two counter tables for comparison? Actually no, because we could increment the counter for each letter in $s$ and decrement the counter for each letter in$t$, then check if the counter reaches back to zero.

public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) {
        return false;
    }
    int[] counter = new int[26];
    for (int i = 0; i < s.length(); i++) {
        counter[s.charAt(i) - 'a']++;           //Returns the char value at the specified index. An index ranges from 0 to length() - 1
        counter[t.charAt(i) - 'a']--;
    }
    for (int count : counter) {
        if (count != 0) {
            return false;
        }
    }
    return true;
}

public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) {
        return false;
    }
    int[] table = new int[26];
    for (int i = 0; i < s.length(); i++) {
        table[s.charAt(i) - 'a']++;
    }
    for (int i = 0; i < t.length(); i++) {
        table[t.charAt(i) - 'a']--;
        if (table[t.charAt(i) - 'a'] < 0) {
            return false;
        }
    }
    return true;
}

Complexity analysis

• Time complexity : $O(n)$. Time complexity is $O(n)$ because accessing the counter table is a constant time operation.

• Space complexity : $O(1)$. Although we do use extra space, the space complexity is $O(1)$ because the table's size stays constant no matter how large $n$ is.