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How can i split a string into words without split function in javascript

How can I split string into words without using split function in javascript. I wonder for a little help.
this is my code:
function trocearCadena(cadena) {
var posEspacios = buscarEspacios(cadena);
var palabras = [];
var j = 0;
while (true) {
var pos = posEspacios.shift();
var subcad = (j == 0) ? cadena.substring(0, pos) : cadena.substring(j + 1, pos);
palabras.push(subcad);
j += pos;
if (j > cadena.length) {
var ultpal = cadena.substring(pos + 1);
palabras.push(ultpal);
break;
}
}
return palabras;
}
function buscarEspacios(cadena) {
var espacios = [];
var pos = -1;
do{
pos = cadena.indexOf(" ", ++pos);
if (pos != -1) espacios.push(pos);
} while (pos != -1);
return espacios;
}

I don't understand what your variable names mean, so I wasn't able to fix the code. Here's another one:
str = "How can I split string into words without using split function in javascript."
var words = [""];
for(var i = 0; i < str.length; i++)
if(str[i] !== " ")
words[words.length - 1] += str[i];
else if(words[words.length - 1])
words.push("");
document.write(words)

Assuming that you also can't use what #Oriol suggested, I would use a recursive function like in the following example:
function _split(s, arr) {
var
str = s.trim(),
words = arr || [],
i = str.indexOf(' ');
if(i !== -1) {
words.push(str.substr(0, i)); // collect the next word
return _split(str.slice(i + 1), words); // recur with new string and words array to keep collecting
} else {
words.push(str); // collect the last word
return words;
}
}
Usage:
_split(' one two three ');
//=> ["one", "two", "three"]

JavaScript procedure to find the characters that are in one string but not in another

I'e been trying to write one and it's getting messy!
Suppose I have two strings textStart, textTarget and I want to keep track of the characters I would need to add and remove from textStart in order to product textTarget.
For instance, if textStart = "dude" and textTarget = "deck", then characters that would need to be added would be 'c' and 'k' and the characters that would need to be substracted would be the 'u' and one of the 'd's.
I'm thinking that I first need to create maps that represent the number of each character in textStart and textTarget.
So I wrote this:
var startChars = {};
for (var k = 0, n = textStart.length; k < n; ++k)
{
if (textStart[k] in startChars)
++startChars[textStart[k]];
else
startChars[textStart[k]] = 1;
}
var targetChars = {};
for (var k = 0, n = textTarget.length; k < n; ++k)
{
if (textTarget[k] in startChars)
++targetChars[textTarget[k]];
else
map1[targetChars[k]] = 1;
}
Which would give me
startChars['d']=2,
startChars['u']=1,
startChars['e']=1
and
targetChars['d']=1,
targetChars['e']=1,
targetChars['c']=1,
targetChars['k']=1
Then I can make create maps needAdded and needRemoved that look at the difference in the above two maps:
var needAdded = {};
var needRemoved = {};
I'm not sure how to fill those maps as intended, because I don't know how to iterate through the keys of a map using JavaScript. I somehow need to end up with
needAdded['c']=1,
needAdded['k']=1,
needRemoved['u']=1,
needRemoved['d']=1
That's where you guys come in and help me.
I hope I've done a good job describing what I'm trying to do and how I've tried to do it so far. My programming intuition tells me that I'm writing too many lines of code and that I need to consult StackOverflow for help. Any way to do this elegantly without JQuery or Regex? I know someone's going to come in this thread and write a 1-line Regex solution or something like that.

var s = 'dude',
t = 'deck',
finalOutput = '';
for (var i = 0; i < s.length; i++){
if ( typeof t[i] != 'undefined' ){
if ( s[i] != t[i] ){
console.log(s[i] + ' changed to ' + t[i]);
s[i] = t[i];
finalOutput += t[i];
} else{
finalOutput += s[i];
}
}
}
console.log('FINAL: ' + finalOutput);

Here's a jsfiddle I just spent way too much time on... hopefully it makes sense :)
var textStart = 'dude';
var textTarget = 'deck';
var startChars = {};
for (var k = 0, n = textStart.length; k < n; ++k)
{
if (textStart[k] in startChars)
++startChars[textStart[k]];
else
startChars[textStart[k]] = 1;
}
var targetChars = {};
for (var k = 0, n = textTarget.length; k < n; ++k)
{
if (textTarget[k] in targetChars)
++targetChars[textTarget[k]];
else
targetChars[textTarget[k]] = 1;
}
console.log('start: ' + JSON.stringify(startChars));
console.log('target: ' + JSON.stringify(targetChars));
var needAdded = {};
var needRemoved = {};
for (var c in startChars) {
// If target does not contain letter, remove all, otherwise remove excess
if (targetChars[c] > 0) {
if (startChars[c] > targetChars[c])
needRemoved[c] = startChars[c] - targetChars[c];
else if (startChars[c] < targetChars[c])
needAdded[c] = targetChars[c] - startChars[c];
} else {
needRemoved[c] = startChars[c];
}
}
for (var c in targetChars) {
// If start does not contain letter, add all, otherwise add excess
if (startChars[c] > 0) {
if (startChars[c] > targetChars[c])
needRemoved[c] = startChars[c] - targetChars[c];
else if (startChars[c] < targetChars[c])
needAdded[c] = targetChars[c] - startChars[c];
} else {
needAdded[c] = targetChars[c];
}
}
console.log('needAdded: ' + JSON.stringify(needAdded));
console.log('needRemoved: ' + JSON.stringify(needRemoved));
The output is as follows:
start: {"d":2,"u":1,"e":1}
target: {"d":1,"e":1,"c":1,"k":1}
needAdded: {"c":1,"k":1}
needRemoved: {"d":1,"u":1}

Ok, also too much time on this:
var textStart = "dude";
var textTarget = "duck";
var map = {};
MapCharacters(textStart, map, 1);
MapCharacters(textTarget, map, -1);
console.log(map);
var toDelete = [];
var toAdd = [];
for (var prop in map) {
if (map.hasOwnProperty(prop)) {
while (map[prop] > 0) {
toDelete.push(prop);
map[prop]--;
}
while (map[prop] < 0) {
toAdd.push(prop);
map[prop]++;
}
}
}
console.log(toDelete);
console.log(toAdd);
function MapCharacters(string, map, add) {
for (var k = 0, n = string.length; k < n; ++k) {
if (string[k] in map) {
map[string[k]] += add;
} else {
map[string[k]] = add;
}
}
}
http://jsfiddle.net/nSV2J/1/
It could probably be done more efficiently, but as I said - too much time!
I realized that the best way to do this is not to make two maps, but just one. In the first case you increment the count for each letter and in the second case you decrease it. Now it's easy to find which ones need to be removed (the ones that end up > 0) and which ones need to be added (the ones that end up < 0)

Showing unique characters in a string only once

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.

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"

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. :-)

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("");
}

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;
});
}

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

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));

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.

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"

<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>

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;
}

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

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"));

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;
};

const countUnique = (s1, s2) => new Set(s1 + s2).size
a shorter way based on #le_m answer

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

JavaScript Decrementing Alphabet Function

New to the JavaScript language and need help creating a function which generates the following display.
abcdefghijklmnopqrstuvwxyz
bcdefghijklmnopqrstuvwxyz
cdefghijklmnopqrstuvwxyz
defghijklmnopqrstuvwxyz
.... and so on, all the way down to
xyz
yz
z
I am not asking for handouts, just a little kickstart for a beginner to get started! Links, hints, tips, anything helps! Thanks!

Arrays and loops are powerful when combined.
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
console.log(alphabet.join(''));
while (alphabet.​length > 0) {
alphabet.shift();
console.log(alphabet.join(''));
}​​​
Edit:
If you really need your decremented alphabet to be left-padded, you can use this:
var alphabet = "abcdefghijklmnopqrstuvwxyz";
var letters = alphabet.split('');
var addPadding = (function (minLength) {
return function (shortString) {
if (shortString.length < minLength) {
return new Array(
minLength - shortString.length + 1
).join(' ') + shortString;
}
};
}(alphabet.length));
console.log(alphabet);
while (letters.length > 0) {
letters.shift();
console.log(addPadding(letters.join('')));
}
Edit:
Here is a much simpler answer:
function decrementingAlphabet () {
var alphabet = 'abcdefghijklmnopqrstuvwxyz';
function iterate(spaces, letters) {
if (letters.length > 0) {
console.log(spaces + letters);
iterate(spaces + ' ', letters.substring(1));
} else {
return;
}
}
iterate('', alphabet);
}

This is simple example.
var str = '';
for (var s=0; s < 26; ++s) {
str = '';
for (var i=0; i < 26 - s; ++i) {
str += String.fromCharCode(97+s+i);
}
document.write(str + "<br/>");
}
​
See http://jsfiddle.net/G5Gds

Hmm, maybe this will help put you on the right track?
var str = '';
// Find out what 'a' is in ASCII
var baseLetterCode = 'a'.charCodeAt(0);
// Loop once for each letter
for (var i = 0; i < 26; ++i) {
// Append to string
str += String.fromCharCode(baseLetterCode + i);
}

In character codes, small alphabets lie from 97 onwards(97 for a). You need to use 2 for loops to print such series.
Here is your jsfiddle demo:
var display='';
for(var i=97;i<123;i++){
var s='';
for(var j=i;j<123;j++){
s+= String.fromCharCode( j );
}
display+=s;
}
alert(display);

(function() {
var theENalphabet = [];
for (var charNow = "a".charCodeAt(0); charNow <= "z".charCodeAt(0); charNow += 1) {
theENalphabet.push(String.fromCharCode(charNow));
}
var isNow = 0;
function decrAlph(startAt) {
var alphString = "";
for (var i = startAt; i < theENalphabet.length; i += 1) {
alphString += theENalphabet[i];
}
console.log(alphString);
isNow++;
while (isNow < theENalphabet.length) {
decrAlph(startAt + 1);
}
};
decrAlph(0);
})();
The charCode getting could be abstracted into a function:
var getCharCode = function(el){
return String.prototype.charCodeAt.call(el, 0);
};
getCharCode("a"); // returns 97..

Find the characters in a string which are not duplicated

I have to make a function in JavaScript that removes all duplicated letters in a string. So far I've been able to do this: If I have the word "anaconda" it shows me as a result "anaconda" when it should show "cod". Here is my code:
function find_unique_characters( string ){
var unique='';
for(var i=0; i<string.length; i++){
if(unique.indexOf(string[i])==-1){
unique += string[i];
}
}
return unique;
}
console.log(find_unique_characters('baraban'));

We can also now clean things up using filter method:
function removeDuplicateCharacters(string) {
return string
.split('')
.filter(function(item, pos, self) {
return self.indexOf(item) == pos;
})
.join('');
}
console.log(removeDuplicateCharacters('baraban'));
Working example:

function find_unique_characters(str) {
var unique = '';
for (var i = 0; i < str.length; i++) {
if (str.lastIndexOf(str[i]) == str.indexOf(str[i])) {
unique += str[i];
}
}
return unique;
}
console.log(find_unique_characters('baraban'));
console.log(find_unique_characters('anaconda'));
If you only want to return characters that appear occur once in a string, check if their last occurrence is at the same position as their first occurrence.
Your code was returning all characters in the string at least once, instead of only returning characters that occur no more than once. but obviously you know that already, otherwise there wouldn't be a question ;-)

Just wanted to add my solution for fun:
function removeDoubles(string) {
var mapping = {};
var newString = '';
for (var i = 0; i < string.length; i++) {
if (!(string[i] in mapping)) {
newString += string[i];
mapping[string[i]] = true;
}
}
return newString;
}

With lodash:
_.uniq('baraban').join(''); // returns 'barn'

You can put character as parameter which want to remove as unique like this
function find_unique_characters(str, char){
return [...new Set(str.split(char))].join(char);
}
function find_unique_characters(str, char){
return [...new Set(str.split(char))].join(char);
}
let result = find_unique_characters("aaaha ok yet?", "a");
console.log(result);

//One simple way to remove redundecy of Char in String
var char = "aaavsvvssff"; //Input string
var rst=char.charAt(0);
for(var i=1;i<char.length;i++){
var isExist = rst.search(char.charAt(i));
isExist >=0 ?0:(rst += char.charAt(i) );
}
console.log(JSON.stringify(rst)); //output string : avsf

For strings (in one line)
removeDuplicatesStr = str => [...new Set(str)].join('');
For arrays (in one line)
removeDuplicatesArr = arr => [...new Set(arr)]

Using Set:
removeDuplicates = str => [...new Set(str)].join('');
Thanks to David comment below.

DEMO
function find_unique_characters( string ){
unique=[];
while(string.length>0){
var char = string.charAt(0);
var re = new RegExp(char,"g");
if (string.match(re).length===1) unique.push(char);
string=string.replace(re,"");
}
return unique.join("");
}
console.log(find_unique_characters('baraban')); // rn
console.log(find_unique_characters('anaconda')); //cod
​

var str = 'anaconda'.split('');
var rmDup = str.filter(function(val, i, str){
return str.lastIndexOf(val) === str.indexOf(val);
});
console.log(rmDup); //prints ["c", "o", "d"]
Please verify here: https://jsfiddle.net/jmgy8eg9/1/

Using Set() and destructuring twice is shorter:
const str = 'aaaaaaaabbbbbbbbbbbbbcdeeeeefggggg';
const unique = [...new Set([...str])].join('');
console.log(unique);

Yet another way to remove all letters that appear more than once:
function find_unique_characters( string ) {
var mapping = {};
for(var i = 0; i < string.length; i++) {
var letter = string[i].toString();
mapping[letter] = mapping[letter] + 1 || 1;
}
var unique = '';
for (var letter in mapping) {
if (mapping[letter] === 1)
unique += letter;
}
return unique;
}
Live test case.
Explanation: you loop once over all the characters in the string, mapping each character to the amount of times it occurred in the string. Then you iterate over the items (letters that appeared in the string) and pick only those which appeared only once.

function removeDup(str) {
var arOut = [];
for (var i=0; i < str.length; i++) {
var c = str.charAt(i);
if (c === '_') continue;
if (str.indexOf(c, i+1) === -1) {
arOut.push(c);
}
else {
var rx = new RegExp(c, "g");
str = str.replace(rx, '_');
}
}
return arOut.join('');
}

I have FF/Chrome, on which this works:
var h={};
"anaconda".split("").
map(function(c){h[c] |= 0; h[c]++; return c}).
filter(function(c){return h[c] == 1}).
join("")
Which you can reuse if you write a function like:
function nonRepeaters(s) {
var h={};
return s.split("").
map(function(c){h[c] |= 0; h[c]++; return c}).
filter(function(c){return h[c] == 1}).
join("");
}
For older browsers that lack map, filter etc, I'm guessing that it could be emulated by jQuery or prototype...

This code worked for me on removing duplicate(repeated) characters from a string (even if its words separated by space)
Link: Working Sample JSFiddle
/* This assumes you have trim the string and checked if it empty */
function RemoveDuplicateChars(str) {
var curr_index = 0;
var curr_char;
var strSplit;
var found_first;
while (curr_char != '') {
curr_char = str.charAt(curr_index);
/* Ignore spaces */
if (curr_char == ' ') {
curr_index++;
continue;
}
strSplit = str.split('');
found_first = false;
for (var i=0;i<strSplit.length;i++) {
if(str.charAt(i) == curr_char && !found_first)
found_first = true;
else if (str.charAt(i) == curr_char && found_first) {
/* Remove it from the string */
str = setCharAt(str,i,'');
}
}
curr_index++;
}
return str;
}
function setCharAt(str,index,chr) {
if(index > str.length-1) return str;
return str.substr(0,index) + chr + str.substr(index+1);
}

Here's what I used - haven't tested it for spaces or special characters, but should work fine for pure strings:
function uniquereduce(instring){
outstring = ''
instringarray = instring.split('')
used = {}
for (var i = 0; i < instringarray.length; i++) {
if(!used[instringarray[i]]){
used[instringarray[i]] = true
outstring += instringarray[i]
}
}
return outstring
}

Just came across a similar issue (finding the duplicates). Essentially, use a hash to keep track of the character occurrence counts, and build a new string with the "one-hit wonders":
function oneHitWonders(input) {
var a = input.split('');
var l = a.length;
var i = 0;
var h = {};
var r = "";
while (i < l) {
h[a[i]] = (h[a[i]] || 0) + 1;
i += 1;
}
for (var c in h) {
if (h[c] === 1) {
r += c;
}
}
return r;
}
Usage:
var a = "anaconda";
var b = oneHitWonders(a); // b === "cod"

Try this code, it works :)
var str="anaconda";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){
return obj;
}
}
).join("");
//output: "cod"

This should work using Regex ;
NOTE: Actually, i dont know how this regex works ,but i knew its 'shorthand' ,
so,i would have Explain to you better about meaning of this /(.+)(?=.*?\1)/g;.
this regex only return to me the duplicated character in an array ,so i looped through it to got the length of the repeated characters .but this does not work for a special characters like "#" "_" "-", but its give you expected result ; including those special characters if any
function removeDuplicates(str){
var REPEATED_CHARS_REGEX = /(.+)(?=.*?\1)/g;
var res = str.match(REPEATED_CHARS_REGEX);
var word = res.slice(0,1);
var raw = res.slice(1);
var together = new String (word+raw);
var fer = together.toString();
var length = fer.length;
// my sorted duplicate;
var result = '';
for(var i = 0; i < str.length; i++) {
if(result.indexOf(str[i]) < 0) {
result += str[i];
}
}
return {uniques: result,duplicates: length};
} removeDuplicates('anaconda')
The regular expression /([a-zA-Z])\1+$/ is looking for:
([a-zA-Z]]) - A letter which it captures in the first group; then
\1+ - immediately following it one or more copies of that letter; then
$ - the end of the string.
Changing it to /([a-zA-Z]).*?\1/ instead searches for:
([a-zA-Z]) - A letter which it captures in the first group; then
.*? - zero or more characters (the ? denotes as few as possible); until
\1 - it finds a repeat of the first matched character.

I have 3 loopless, one-line approaches to this.
Approach 1 - removes duplicates, and preserves original character order:
var str = "anaconda";
var newstr = str.replace(new RegExp("[^"+str.split("").sort().join("").replace(/(.)\1+/g, "").replace(/[.?*+^$[\]\\(){}|-]/g, "\\$&")+"]","g"),"");
//cod
Approach 2 - removes duplicates but does NOT preserve character order, but may be faster than Approach 1 because it uses less Regular Expressions:
var str = "anaconda";
var newstr = str.split("").sort().join("").replace(/(.)\1+/g, "");
//cdo
Approach 3 - removes duplicates, but keeps the unique values (also does not preserve character order):
var str = "anaconda";
var newstr = str.split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
//acdno

function removeduplicate(str) {
let map = new Map();
// n
for (let i = 0; i < str.length; i++) {
if (map.has(str[i])) {
map.set(str[i], map.get(str[i]) + 1);
} else {
map.set(str[i], 1);
}
}
let res = '';
for (let i = 0; i < str.length; i++) {
if (map.get(str[i]) === 1) {
res += str[i];
}
}
// o (2n) - > O(n)
// space o(n)
return res;
}

If you want your function to just return you a unique set of characters in your argument, this piece of code might come in handy.
Here, you can also check for non-unique values which are being recorded in 'nonUnique' titled array:
function remDups(str){
if(!str.length)
return '';
var obj = {};
var unique = [];
var notUnique = [];
for(var i = 0; i < str.length; i++){
obj[str[i]] = (obj[str[i]] || 0) + 1;
}
Object.keys(obj).filter(function(el,ind){
if(obj[el] === 1){
unique+=el;
}
else if(obj[el] > 1){
notUnique+=el;
}
});
return unique;
}
console.log(remDups('anaconda')); //prints 'cod'
If you want to return the set of characters with their just one-time occurrences in the passed string, following piece of code might come in handy:
function remDups(str){
if(!str.length)
return '';
var s = str.split('');
var obj = {};
for(var i = 0; i < s.length; i++){
obj[s[i]] = (obj[s[i]] || 0) + 1;
}
return Object.keys(obj).join('');
}
console.log(remDups('anaconda')); //prints 'ancod'

function removeDuplicates(str) {
var result = "";
var freq = {};
for(i=0;i<str.length;i++){
let char = str[i];
if(freq[char]) {
freq[char]++;
} else {
freq[char] =1
result +=char;
}
}
return result;
}

console.log(("anaconda").split('').sort().join('').replace(/(.)\1+/g, ""));
By this, you can do it in one line.
output: 'cdo'

function removeDuplicates(string){
return string.split('').filter((item, pos, self)=> self.indexOf(item) == pos).join('');
}
the filter will remove all characters has seen before using the index of item and position of the current element

Method 1 : one Simple way with just includes JS- function
var data = 'sssssddddddddddfffffff';
var ary = [];
var item = '';
for (const index in data) {
if (!ary.includes(data[index])) {
ary[index] = data[index];
item += data[index];
}
}
console.log(item);
Method 2 : Yes we can make this possible without using JavaScript function :
var name = 'sssssddddddddddfffffff';
let i = 0;
let newarry = [];
for (let singlestr of name) {
newarry[i] = singlestr;
i++;
}
// now we have new Array and length of string
length = i;
function getLocation(recArray, item, arrayLength) {
firstLaction = -1;
for (let i = 0; i < arrayLength; i++) {
if (recArray[i] === item) {
firstLaction = i;
break;
}
}
return firstLaction;
}
let finalString = '';
for (let b = 0; b < length; b++) {
const result = getLocation(newarry, newarry[b], length);
if (result === b) {
finalString += newarry[b];
}
}
console.log(finalString); // sdf

// Try this way
const str = 'anaconda';
const printUniqueChar = str => {
const strArr = str.split("");
const uniqueArray = strArr.filter(el => {
return strArr.indexOf(el) === strArr.lastIndexOf(el);
});
return uniqueArray.join("");
};
console.log(printUniqueChar(str)); // output-> cod

function RemDuplchar(str)
{
var index={},uniq='',i=0;
while(i<str.length)
{
if (!index[str[i]])
{
index[str[i]]=true;
uniq=uniq+str[i];
}
i++;
}
return uniq;
}

We can remove the duplicate or similar elements in string using for loop and extracting string methods like slice, substring, substr
Example if you want to remove duplicate elements such as aababbafabbb:
var data = document.getElementById("id").value
for(var i = 0; i < data.length; i++)
{
for(var j = i + 1; j < data.length; j++)
{
if(data.charAt(i)==data.charAt(j))
{
data = data.substring(0, j) + data.substring(j + 1);
j = j - 1;
console.log(data);
}
}
}
Please let me know if you want some additional information.