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LeetCode link: 242. Valid Anagram, difficulty: Easy.

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.

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

Output: true

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

Output: false

• 1 <= s.length, t.length <= 5 * 10^4
• s and t consist of lowercase English letters.

1. If the lengths of the two strings are different, return false directly.
2. Use two hash tables to store the statistics of the two strings respectively, with the key being the character and the value being the number of characters.
3. Compare the two hash tables to see if they are equal.

• Time complexity: O(N).
• Space complexity: O(N).

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        var sCharToCount = new HashMap<Character, Integer>();
        for (var character : s.toCharArray()) {
            sCharToCount.put(character, sCharToCount.getOrDefault(character, 0) + 1);
        }

        var tCharToCount = new HashMap<Character, Integer>();
        for (var character : t.toCharArray()) {
            tCharToCount.put(character, tCharToCount.getOrDefault(character, 0) + 1);
        }

        return sCharToCount.equals(tCharToCount);
    }
}

# from collections import defaultdict

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False

        s_char_to_count = defaultdict(int)
        for char in s:
            s_char_to_count[char] += 1

        t_char_to_count = defaultdict(int)
        for char in t:
            t_char_to_count[char] += 1

        return s_char_to_count == t_char_to_count

var isAnagram = function (s, t) {
  if (s.length != t.length) {
    return false;
  }

  const sCharToCount = new Map()
  const tCharToCount = new Map()

  for (const char of s) {
    sCharToCount.set(char, (sCharToCount.get(char) || 0) + 1)
  }

  for (const char of t) {
    tCharToCount.set(char, (tCharToCount.get(char) || 0) + 1)
  }

  return _.isEqual(sCharToCount, tCharToCount)
};

public class Solution
{
    public bool IsAnagram(string s, string t)
    {
        if (s.Length != t.Length)
            return false;

        var sCharToCount = new Dictionary<char, int>();
        var tCharToCount = new Dictionary<char, int>();

        foreach (char character in s)
            sCharToCount[character] = sCharToCount.GetValueOrDefault(character, 0) + 1;

        foreach (char character in t)
            tCharToCount[character] = tCharToCount.GetValueOrDefault(character, 0) + 1;

        foreach (var entry in sCharToCount)
        {
            if (entry.Value != tCharToCount.GetValueOrDefault(entry.Key))
            {
                return false;
            }
        }

        return true;
    }
}

import "reflect"

func isAnagram(s string, t string) bool {
    if len(s) != len(t) {
        return false
    }

    // Create frequency maps for both strings
    sCharCount := make(map[rune]int)
    for _, char := range s {
        sCharCount[char]++
    }

    tCharCount := make(map[rune]int)
    for _, char := range t {
        tCharCount[char]++
    }

    return reflect.DeepEqual(sCharCount, tCharCount)
}

def is_anagram(s, t)
  return false if s.length != t.length

  s_char_to_count = Hash.new(0)
  t_char_to_count = Hash.new(0)

  s.each_char do |char|
    s_char_to_count[char] += 1
  end

  t.each_char do |char|
    t_char_to_count[char] += 1
  end

  s_char_to_count == t_char_to_count
end

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }

        unordered_map<char, int> s_char_to_count;
        for (char character : s) {
            s_char_to_count[character]++;
        }

        unordered_map<char, int> t_char_to_count;
        for (char character : t) {
            t_char_to_count[character]++;
        }

        return s_char_to_count == t_char_to_count;
    }
};

// Welcome to create a PR to complete the code of this language, thanks!

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Original link: 242. Valid Anagram - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions.

GitHub repository: leetcode-python-java.