# 242. Valid Anagram
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
Approach 1: Sorting.
Approach 2: Hash table.
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
class Solution {
public:
bool isAnagram(string s, string t) {
if (s.size() != t.size())
return false;
std::sort(s.begin(), s.end());
std::sort(t.begin(), t.end());
return s == t;
}
};
Approach 2:
class Solution {
public:
bool isAnagram(string s, string t) {
if (s.size() != t.size())
return false;
unordered_map<char,int> visited;
for (int i = 0; i < s.size(); ++i)
{
visited[s[i]]++;
}
for (int i = 0; i < t.size(); ++i)
{
/*
if (visited.find(t[i]) == visited.end())
return false;
visited[t[i]]--;
if (visited[t[i]] == 0)
visited.erase(t[i]);
*/
visited[t[i]]--;
if (visited[t[i]] < 0)
return false;
}
return true;
}
};
class Solution {
public:
bool isAnagram(string s, string t) {
if (s.size() != t.size())
return false;
unordered_map<char,int> visited;
for (int i = 0; i < s.size(); ++i)
{
visited[s[i]]++;
visited[t[i]]--;
}
for (auto it = visited.begin(); it != visited.end(); ++it)
{
if (it->second < 0)
return false;
}
return true;
}
};