Valid Anagram (easy)

Problem Statement

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.

Example 1:

Input: s = "listen", t = "silent"
Output: true

Example 2:

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

Example 3:

Input: s = "hello", t = "world"
Output: false

Constraints:

1 <= s.length, t.length <= 5 * 10<sup>4</sup>
s and t consist of lowercase English letters.

.....

Like the course? Get enrolled and start learning!

Ejike Nwude

· 3 years ago

Your testcases allow the following solution to pass when it should fail. For instance the following are not anagrams {add, ada} but since a Set does not account for the frequency, it passes them as anagrams:

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

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

        Set<Character> characters = new HashSet<>();
        for (char c : s.toCharArray()) {
            characters.add(c);
        }

        for (char character : t.toCharArray()) {
            if (!characters.contains(character)) {
                return false;
            }
        }
        return true;
    }
}

Show 1 reply

Made Doddy Adi Pranatha

· 3 years ago

function isAnagram(s, t) {
    // TODO: Write your code here
    return s.split("").sort().join("") === t.split("").sort().join("");
}

Thai Minh

· 3 years ago

class Solution:
  def isAnagram(self, s, t):
    # TODO: Write your code here
    # create dict for string s
    # check char of t for s dict if it is not in s then false
    s_dict = {}
    for ch in s:
      s_dict[ch] = s_dict.get(ch, 0) + 1
    for ch in t:
      if ch not in s: 
        return False
      elif ch in s and s_dict[ch] - 1 < 0:
        return False
      else:
        s_dict[ch] -= 1
    return True

amazonintern101

· 3 years ago

class Solution:
  def isAnagram(self, s, t):
    # TODO: Write your code here
    charSet = {}

    if len(s) != len(t):
      return False

    for char in s:
      if char not in charSet:
        charSet[char] = 0
      else:
        charSet[char] += 1
    
    for char in t:
      if char not in charSet:
        return False
      else:
        if charSet[char] >= 0:
          charSet[char] -= 1
        else:
          return False

    return True

Alex

· 3 years ago

Show 1 reply

hassan.javeed84

· 2 years ago

This code passes all the tests however it's wrong as in case of s = "hel" and t = "hll" it returns True

class Solution:
  def isAnagram(self, s, t):
    # TODO: Write your code here
    if len(s) != len(t):
      return False
    
    l = list(s)

    for c in t:
      if c not in l:
        return False

    return True

Show 1 reply

Amber Wolf

· a year ago

This was my solution to the question. There is however one test case in the web browser IDE that is incorrect. In my solution I accounted for a string that can be given that is empty, which by definition is not an anagram. So a string "" being compared to any other string, including "" should be false as that is not an anagram.

public bool isAnagram(string s, string t)
{
    // using Regex to massage the strings and remove any special characters/white spaces
    var trimmedS = Regex.Replace(s, "[^a-zA-Z0-9]", "").ToLower();
    var trimmedT = Regex.Replace(t, "[^a-zA-Z0-9]", "").ToLower();
    // first checking the length of the words, if they are a different amount of charcters or they contain no characters they cannot be an anagram.
    if (trimmedS.Length != trimmedT.Length

Bruno Raiado

· 4 months ago

using System;
using System.Collections.Generic;

public class Solution {
    public bool isAnagram(string s, string t) {
      if(s.Length != t.Length) return false;
      int asciiS = 0;
      int asciiT = 0;
      for(int i =0;i<s.Length;i++) {
        asciiS += (int)s[i];
        asciiT += (int)t[i];
      }
      return asciiT == asciiS;
    }
}

Carol Lisbon

· 4 months ago

class Solution:
  def isAnagram(self, s, t):
    if len(s) != len(t):
      return False
    
    s_dict = dict()
    t_dict = dict()

    for s1, t1 in zip(s,t):
      if s1 not in s_dict:
        s_dict[s1] = 0
      if t1 not in t_dict:
        t_dict[t1] = 0
      s_dict[s1] += 1
      t_dict[t1] += 1

    return s_dict == t_dict

xha80n+9p6ne

· 3 months ago

from collections import Counter 

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

    return Counter(s) == Counter(t)