Grokking the Coding Interview: Patterns for Coding Questions
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Valid Anagram (easy)

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A

Anonymous

· 3 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
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A

Anonymous

· 3 years ago

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

Anonymous

· 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;
    }
}
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A

Anonymous

· 3 years ago

By definition an anagram has to be a word or set of words that can be re-arranged into another word or set of words. An empty string isn't a word and can't be rearranged

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A

Anonymous

· 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
A

Anonymous

· 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

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