leetcode:242. Valid Letter ectopic words

242. Effective alphabetic words

source ： Power button （LeetCode）

link : https://leetcode.cn/problems/valid-anagram/

Given two strings s and t , Write a function to determine t Whether it is s Letter heteronym of .
Be careful ： if s and t Each character in the has the same number of occurrences , said s and t They are mutually alphabetic words .

Example 1:

 Input : s = "anagram", t = "nagaram"
 Output : true

Example 2:

 Input : s = "rat", t = "car"
 Output : false

Tips :

• 1 <= s.length, t.length <= $5\ast 10^4$
• s and t Only lowercase letters

Advanced : If the input string contains unicode What about characters ？ Can you adjust your solution to deal with this situation ？

solution

• Sort ： The two are sorted and then compared element by element ;
• Hashtable ： First, judge whether the length of the two is the same , Return directly if the length is different false; In addition, a hash table is used to traverse the structure of comparison storage ,key Is an element ,value Is the number of times the element appears ; Then iterate over another string , If the element is in the hash table , Reduce the number of times 1, If you reduce it to a negative number , Go straight back to false, If not in the hash table , And go straight back to false; Traverse to the end , Look at each... In the hash table key Whether the corresponding times are 0;

Code implementation

Sort

Use the properties of the list ;
python Realization

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        return sorted(s) == sorted(t)

c++ Realization

class Solution {
    
public:
    bool isAnagram(string s, string t) {
    
        if (s.size() != t.size())
            return false;
        
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());

        for (int i=0; i<s.size(); i++) {
    
            if (s[i] != t[i])
                return false;
        }
        return true;
    }
};

Complexity analysis

• Time complexity ： $O(NlogN)$
• Spatial complexity ： $O(logN)$ Sorting requires space

Hashtable

python Realization

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        
        counts = dict()
        for ch in s:
            if ch in counts:
                counts[ch] += 1
            else:
                counts[ch] = 1
        
        for ch in t:
            if ch in counts:
                counts[ch] -= 1
                if counts[ch] == 0:
                    del counts[ch]  #  The deletion times are 0 Of key 
            else:
                return False
        
        return len(counts) == 0

c++ Realization

class Solution {
    
public:
    bool isAnagram(string s, string t) {
    
        if (s.size() != t.size())
            return false;
        
        unordered_map<char, int> counts;

        for(char ch: s) {
    
            auto it = counts.find(ch);
            if (it != counts.end())
                counts[ch]++;
            else
                counts[ch] = 1;
        }

        for (char ch: t) {
    
            auto it = counts.find(ch);
            if (it == counts.end())
                return false;
            counts[ch]--;
            if (counts[ch] == 0)
                counts.erase(ch);
        }

        return counts.size() == 0;
    }
};

Complexity analysis

• Time complexity ： $O(N)$
• Spatial complexity ： $O(S)$ The number of characters