242. Valid Anagram

hashmap O(n) time O(1) space

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) return false;
        int f[256] = {0};
        for (char c : s) {
            ++f[c];
        }
        for (char c : t) {
            if (--f[c] < 0) return false;
        }
        return true;
    }
};

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (size(s) != size(t)) return false;
        int f[256] = {0};
        for (int i = 0; i < size(s); ++i) {
            ++f[s[i]];
            --f[t[i]];
        }
        return all_of(f, f + 256, [](int x){ return x == 0; });
    }
};

Follow up

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.