Javascript内部数组长度中的双循环

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: "(((("

Answer 1

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: "(((("

Answer 2

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")); // ((((

Answer 3

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: "(((("

Answer 4

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);

Answer 5

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")); // ((((