CODEWITHSUNDEEP
javascriptregexperformance
GreatGaja
GreatGaja

Reputation: 405

Finding characters of a word in a string, optimized

I am doing doing a few code challenges in the hope to learn some new stuff. Currently I have written a piece of code that find charachter of a given word in a string of random letters.

I thought regexp might be the best performance-wise (it's one of the objectives). This code passes the checks but takes too long with absurd long strings. Is there any way I can improve this? It's really ugly as is honestly. I've tried several approaches but I am probably just really a newbie at reg exp etc.

before all the if statements I only used regexp but if str2 which is the word I am looking for had double characters it would come back 'true' because it would count already counted characters. That is why I am using replace to exclude them. That's all I could get.

the goal is to return true if a portion of str1 can be rearranged to form str2, otherwise return false. Only lower case letters will be used (a-z). No punctuation or digits will be included. for example scramble('aabbcamaomsccdd','commas') should return true


function scramble (str1, str2)
{
var o = 0; // tracks amount of matched letters.
 for(i = 0; i < str2.length; i++)
 {
    var regex1 = new RegExp (str2[i]) ; // select letter from word that needs to be found
    if( regex1.test(str1) == true)// if selected character is found us replace to remove it from the random characters string for next iteration.
    {
      str1 = str1.replace(regex1 ,"");
      o++; // increment o if character is removed from random string.
    }
 }
//check if amount of removed characters equals total characters of word that we want.
    if ( o == str2.length)
    {
      return true
    }
    if (o !== str2.length)
    {
      return false
    }
}

Update: I flagged the hash table as answer because afaik this was not doable with regexp it seems also I was able to achieve a proper result with .split and loops myself plus the hash table also achieved this.

Upvotes: 1

Views: 253

Answers (2)

Nina Scholz
Nina Scholz

Reputation: 386604

You could take a hash table, count the wanted characters and return if no count is not more necessary.

function scramble (str1, str2) {
    var counter = {},
        keys = 0;

    for (let i = 0; i < str2.length; i++) {
        if (!counter[str2[i]]) {
            counter[str2[i]] = 0;
            keys++;
        }
        counter[str2[i]]++;
    }

    for (let i = 0; i < str1.length; i++) {
        if (!counter[str1[i]]) continue;
        if (!--counter[str1[i]] && !--keys) return true;
    }
    return false;
}

console.log(scramble('abc', 'cba'));
console.log(scramble('abc', 'aba'));
console.log(scramble('abcdea', 'aba'));
console.log(scramble('aabbcamaomsccdd', 'commas'));

Upvotes: 1

Samuel Gon&#231;alves
Samuel Gon&#231;alves

Reputation: 21

if-less methodology!

i didnt stress it on tests but looks fine

function scramble(str1, str2) {
  str1 = [...str1];
  return [...str2].filter((str => (str == str1.splice(str1.indexOf(str), 1)))).join('') == str2;

}

Upvotes: 1

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