Double for loop in Javascript inner array length

Question

I am trying to create a function that takes in a string and changes each letters value to a "(" if the character is not duplicated in the string, and a ")" if the character does have a duplicate present in the string. I have decided to go an unconventional route to solve this problem but I am running in to an issue with a double for loop. From what I understand, the inner for loop in javascript does not have access to the variables outside of the loop. I want to loop through every item in an array twice but I'm not sure what to set the inner loops length as.

Here is my code:

function sortAndChange(word) {
const splitter = word.toLowerCase().split("");
//let jSplitter = word.toLowerCase().split("").length;
let endResult = "";
let truthArray = [];

for(i = 0; i < splitter.length; i++){
    for(j = 0; j < splitter.length; j++){
        console.log(j);
        if(splitter[i] == splitter[j]){
            truthArray.push(true);
        } else {
            truthArray.push(false);
        }
    }
    console.log(truthArray);
    truthArray.every(item => item === false) ? endResult += "(" : endResult += ")";
    truthArray = [];
}
console.log(endResult);
}

Expected Result:

sortAndChange("Success") //expected output: ")())())"
sortAndChange("easy") //expected output: "(((("

"the inner for loop in javascript does not have access to the variables outside of the loop" that's incorrect. — Aaron
– Aaron, Commented Feb 5, 2020 at 16:47
just noting, in both of those loops you only ever increment i — austin wernli
– austin wernli, Commented Feb 5, 2020 at 16:49

Maheer Ali · Accepted Answer · 2020-02-05 16:54:13Z

3

You can do that in following steps:

Convert string to array using split and use map() on it.
Compare the indexOf() and lastIndexOf() to check if its duplicate or not.
Return the ) or ( based on ur condition. And then at last join the array

function sortAndChange(str){
  let arr = str.toLowerCase().split('')
  return arr.map(x => {
    //if its not duplicated
    if(arr.indexOf(x) === arr.lastIndexOf(x)){
      return '('
    }
    //If its duplicated
    else{
      return ')'
    }
  }).join('');
}

console.log(sortAndChange("Success")) //expected output: ")())())"
console.log(sortAndChange("easy")) //expected output: "(((("

answered Feb 5, 2020 at 16:54

Maheer Ali

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1 Comment

Maheer Ali Over a year ago

@klaurtar1 First you loop through all the values in map and then again in indexOf and lastIndexOf so its O(n ^ 2). If you want linear time-complexity I could think of it.

Nina Scholz · Accepted Answer · 2020-02-05 17:43:36Z

1

You could take a object and keep a boolean value for later mapping the values.

This approach has two loops with O(2n)

function sortAndChange(word) {
    word = word.toLowerCase();
    var map = [...word].reduce((m, c) => (m[c] = c in m, m), {});
    return Array
        .from(word, c => '()'[+map[c]])
        .join('');
}

console.log(sortAndChange("Success")); // )())())
console.log(sortAndChange("easy"));    // ((((

answered Feb 5, 2020 at 17:43

Nina Scholz

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Comments

Titulum · Accepted Answer · 2020-02-05 16:56:49Z

0

This can easily be achieved using a combination of regex and the map construct in javascript:

const input = "this is a test";
const characters = input.toLowerCase().split('');
const transformed = characters.map(currentCharacter => {
  const regexpression = new RegExp(currentCharacter, "g");
  if (input.toLowerCase().match(regexpression || []).length > 1) return ')'
  return '(';
}).join("");
console.log(transformed);

answered Feb 5, 2020 at 16:56

Titulum

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Comments

Orelsanpls · Accepted Answer · 2020-02-05 17:06:19Z

0

Look at the following snippet and comments

function sortAndChange(str) {
  // we create an array containing the characters on the string
  // so we can use Array.reduce
  return str.split('').reduce((tmp, x, xi) => {
    // we look if the character is duplicate in the string
    // by looking for instance of the character
    if (str.slice(xi + 1).includes(x.toLowerCase())) {
      // Duplicate - we replace every occurence of the character
      tmp = tmp.replace(new RegExp(x, 'gi'), ')');
    } else {
      // Not duplicate
      tmp = tmp.replace(new RegExp(x, 'gi'), '(');
    }

    return tmp;
  }, str);
}

console.log(sortAndChange('Success')); //expected output: ")())())"
console.log(sortAndChange('Easy')); //expected output: "(((("

edited Feb 5, 2020 at 17:06

answered Feb 5, 2020 at 16:54

Orelsanpls

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2 Comments

Maheer Ali Over a year ago

Your output is wrong. I the first letter is not duplicated to it should be replaced by (

Orelsanpls Over a year ago

@MaheerAli Fixed it. Had my characters wrong with lowercase. thanks for the input

Siva Kondapi Venkata · Accepted Answer · 2020-02-05 17:42:21Z

0

1) use Array.from to convert to array of chars
2) use reduce to build object with key-value pairs as char in string and ( or ) as value based on repetition .
3) Now convert original string to result string using the chars from above object.

function sortAndChange(str) {
  const str_arr = Array.from(str.toLowerCase());
  const obj = str_arr.reduce(
    (acc, char) => ((acc[char] = char in acc ? ")" : "("), acc),
    {}
  );
  return str_arr.reduce((acc, char) => `${acc}${obj[char]}`, "");
}

console.log(sortAndChange("Success")); // ")())())"
console.log(sortAndChange("easy")); // ((((

answered Feb 5, 2020 at 17:42

Siva Kondapi Venkata

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