Showing unique characters in a string only once

Question

I have a string with repeated letters. I want letters that are repeated more than once to show only once.

Example input: aaabbbccc
Expected output: abc

I've tried to create the code myself, but so far my function has the following problems:

• if the letter doesn't repeat, it's not shown (it should be)
• if it's repeated once, it's show only once (i.e. aa shows a - correct)
• if it's repeated twice, shows all (i.e. aaa shows aaa - should be a)
• if it's repeated 3 times, it shows 6 (if aaaa it shows aaaaaa - should be a)

function unique_char(string) {
    var unique = '';
    var count = 0;
    for (var i = 0; i < string.length; i++) {
        for (var j = i+1; j < string.length; j++) {
            if (string[i] == string[j]) {
                count++;
                unique += string[i];
            }
        }
    }
    return unique;
}

document.write(unique_char('aaabbbccc'));

The function must be with loop inside a loop; that's why the second for is inside the first.

Answer 1

Fill a Set with the characters and concatenate its unique entries:

function unique(str) {
  return String.prototype.concat.call(...new Set(str));
}

console.log(unique('abc'));    // "abc"
console.log(unique('abcabc')); // "abc"

Answer 2

Convert it to an array first, then use Josh Mc’s answer at How to get unique values in an array, and rejoin, like so:

var nonUnique = "ababdefegg";
var unique = Array.from(nonUnique).filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');

All in one line. :-)

Answer 3

Too late may be but still my version of answer to this post:

function extractUniqCharacters(str){
    var temp = {};
    for(var oindex=0;oindex<str.length;oindex++){
        temp[str.charAt(oindex)] = 0; //Assign any value
    }
    return Object.keys(temp).join("");
}

Answer 4

You can use a regular expression with a custom replacement function:

function unique_char(string) {
    return string.replace(/(.)\1*/g, function(sequence, char) {
         if (sequence.length == 1) // if the letter doesn't repeat
             return ""; // its not shown
         if (sequence.length == 2) // if its repeated once
             return char; // its show only once (if aa shows a)
         if (sequence.length == 3) // if its repeated twice
             return sequence; // shows all(if aaa shows aaa)
         if (sequence.length == 4) // if its repeated 3 times
             return Array(7).join(char); // it shows 6( if aaaa shows aaaaaa)
         // else ???
         return sequence;
    });
}

Answer 5

Using lodash:

_.uniq('aaabbbccc').join(''); // gives 'abc'

Answer 6

Per the actual question: "if the letter doesn't repeat its not shown"

function unique_char(str)
{
    var obj = new Object();

    for (var i = 0; i < str.length; i++)
    {
        var chr = str[i];
        if (chr in obj)
        {
            obj[chr] += 1;
        }
        else
        {
            obj[chr] = 1;
        }
    }

    var multiples = [];
    for (key in obj)
    {
        // Remove this test if you just want unique chars
        // But still keep the multiples.push(key)
        if (obj[key] > 1)
        {
            multiples.push(key);
        }
    }

    return multiples.join("");
}

var str = "aaabbbccc";
document.write(unique_char(str));

Answer 7

Your problem is that you are adding to unique every time you find the character in string. Really you should probably do something like this (since you specified the answer must be a nested for loop):

function unique_char(string){

    var str_length=string.length;
    var unique='';

    for(var i=0; i<str_length; i++){

        var foundIt = false;
        for(var j=0; j<unique.length; j++){

            if(string[i]==unique[j]){

                foundIt = true;
                break;
            }

        }

        if(!foundIt){
            unique+=string[i];
        }

    }

   return unique;
}

document.write( unique_char('aaabbbccc'))

In this we only add the character found in string to unique if it isn't already there. This is really not an efficient way to do this at all ... but based on your requirements it should work.

I can't run this since I don't have anything handy to run JavaScript in ... but the theory in this method should work.

Answer 8

Try this if duplicate characters have to be displayed once, i.e., for i/p: aaabbbccc o/p: abc

var str="aaabbbccc";
Array.prototype.map.call(str, 
  (obj,i)=>{
    if(str.indexOf(obj,i+1)==-1 ){
     return obj;
    }
  }
).join("");
//output: "abc"

And try this if only unique characters(String Bombarding Algo) have to be displayed, add another "and" condition to remove the characters which came more than once and display only unique characters, i.e., for i/p: aabbbkaha o/p: kh

var str="aabbbkaha";
Array.prototype.map.call(str, 
 (obj,i)=>{
   if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){ // another and condition
     return obj;
   }
 }
).join("");
//output: "kh"

Answer 9

<script>
    uniqueString = "";

    alert("Displays the number of a specific character in user entered string and then finds the number of unique characters:");

    function countChar(testString, lookFor) {
        var charCounter = 0;
        document.write("Looking at this string:<br>");

        for (pos = 0; pos < testString.length; pos++) {
            if (testString.charAt(pos) == lookFor) {
                charCounter += 1;
                document.write("<B>" + lookFor + "</B>");
            } else
                document.write(testString.charAt(pos));
        }
        document.write("<br><br>");
        return charCounter;
    }

    function findNumberOfUniqueChar(testString) {
        var numChar = 0,
            uniqueChar = 0;
        for (pos = 0; pos < testString.length; pos++) {
            var newLookFor = "";
            for (pos2 = 0; pos2 <= pos; pos2++) {
                if (testString.charAt(pos) == testString.charAt(pos2)) {
                    numChar += 1;
                }
            }
            if (numChar == 1) {
                uniqueChar += 1;
                uniqueString = uniqueString + " " + testString.charAt(pos)
            }
            numChar = 0;
        }
        return uniqueChar;
    }

    var testString = prompt("Give me a string of characters to check", "");
    var lookFor = "startvalue";
    while (lookFor.length > 1) {
        if (lookFor != "startvalue")
            alert("Please select only one character");
        lookFor = prompt(testString + "\n\nWhat should character should I look for?", "");
    }
    document.write("I found " + countChar(testString, lookFor) + " of the<b> " + lookFor + "</B> character");
    document.write("<br><br>I counted the following " + findNumberOfUniqueChar(testString) + " unique character(s):");
    document.write("<br>" + uniqueString)
</script>

Answer 10

Here is the simplest function to do that

  function remove(text) 
    {
      var unique= "";
      for(var i = 0; i < text.length; i++)
      {
        if(unique.indexOf(text.charAt(i)) < 0) 
        {
          unique += text.charAt(i);
        }
      }
      return unique;
    }

Answer 11

The one line solution will be to use Set. const chars = [...new Set(s.split(''))];

Answer 12

If you want to return values in an array, you can use this function below.

const getUniqueChar = (str) => Array.from(str)
    .filter((item, index, arr) => arr.slice(index + 1).indexOf(item) === -1);

console.log(getUniqueChar("aaabbbccc"));

Alternatively, you can use the Set constructor.

const getUniqueChar = (str) => new Set(str);

console.log(getUniqueChar("aaabbbccc"));

Answer 13

Here is the simplest function to do that pt. 2

const showUniqChars = (text) => {
  let uniqChars = "";

  for (const char of text) {
    if (!uniqChars.includes(char))
      uniqChars += char;
  }
  return uniqChars;
};

Answer 14

const countUnique = (s1, s2) => new Set(s1 + s2).size

a shorter way based on @le_m answer

Answer 15

let unique=myArray.filter((item,index,array)=>array.indexOf(item)===index)