Reputation: 221

Why is this anagram function not correct?

Just got back my test for my Javascript class and I missed a question that I felt quite certain about.

We were supposed to make a function that returned true if two strings were anagrams, and false otherwise. I cant find an instance where this wouldn't work. I would appreciate any help!

This is what I had:

function isAnagram(str1, str2){
    if(str1.length !== str2.length){
        return false;
    }
    else{
        for(var i = 0; i < str1.length; i++){
            if(!str2.includes(str1[i])){
                return false;
            }
         }
         return true;
     }
}

Upvotes: 1

Answers (7)

hitesh.gawhade

Reputation: 347

Simply compare both the strings character by character and register how many matches you find. The total matches must be >= to the string length (to cater to the cases where the characters might be repeating e.g. boot)

var i, j;
var z = 0;
function checkAnagram(str1, str2) {
    if (str1.length != str2.length) { return `${str1} and ${str2} are not anagram, lengths dont match!`; }
    else {
        for (i = 0; i < str1.length; i++) {
            for (j = 0; j < str2.length; j++) {
                if(str1[i] == str2[j]) {
               // console.log('matches',str1[i],str2[j]);
                z++;
                }
                }
            }
          //  console.log(z);
            if (z>=str1.length)
            return`${str1} and ${str2} are anagram!`;
            else return `${str1} and ${str2} are not anagram!`;
        }
}
console.log(checkAnagram('fired','fried'));
console.log(checkAnagram('logo','ogol'));
console.log(checkAnagram('abcd','abc'));

Upvotes: 0

Stephan

Reputation: 2115

Single characters can occur multiple times in your function, for example comparing test and tttt returns true. One way to modify your function is to remove the character from str2 each time one is found in the for loop:

function isAnagram(str1, str2){
  if(str1.length !== str2.length) return false;
  for(var i = 0; i < str1.length; i++){
    var found = str2.indexOf(str1[i]);
    if(found === -1) return false;
    str2 = str2.substr(0,found) + str2.substr(found + 1)
  }
  return true;
}

Upvotes: 1

Aron

Reputation: 9246

It only checks that each letter of str1 is present in str2, regardless of how many times they appear.

EDIT: It also doesn't check if str2 has letters that don't appear in str1.

function isAnagram(str1, str2) {
  if (str1.length !== str2.length) {
    return false;
  } else {
    for (var i = 0; i < str1.length; i++) {
      if (!str2.includes(str1[i])) {
        return false;
      }
    }
    return true;
  }
}

console.log(isAnagram('hello', 'hlleo'), ', should be true'); // true: correct
console.log(isAnagram('bye', 'byee'), ', should be false'); // false: correct
console.log(isAnagram('byeb', 'byee'), ', should be false'); // true: incorrect
console.log(isAnagram('byeb', 'byte'), ', should be false'); // true: incorrect

Upvotes: 2

kind user

Reputation: 41893

Alternative solution.

const isAnagram = (s1, s2) => {
  return lowerify(s1) == lowerify(s2);
}

const lowerify = str => str.toLowerCase().split('').sort().join('');

console.log(isAnagram("Arrigo Boito", "Tobia Gorrio"));
console.log(isAnagram("abc", "CBa"));
console.log(isAnagram("deep", "depp"));

Upvotes: 3

Timothy Groote

Reputation: 8653

as a test case, your code would return success on this, but violate the rules of being an anagram :

isAnagram("ooaooa", "oooooa");

it just checks if the same amount of letters are present, and if every letter from word a is present in word b, but not the count of the letters in both words.

Upvotes: 1