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Missing letters freecodecamp

Actually I found an answer a few minutes ago.
But I found something strange.
This is my answer for 'Missing letters' in freeCodeCamp challenges.
function fearNotLetter(str) {
var string;
for (i=0;i<str.length;i++) {
if(str.charCodeAt(i)+1 < str.charCodeAt(i+1)){
string = String.fromCharCode(str.charCodeAt(i)+1);
}
}
return string;
}
When I change < operator in if statement into != (not same), it doesn't work!
For me, it seems that != works exactly same as < operator does.
(Because 'not same' can mean something is bigger than the other.)
What is the difference between < and != in the code above?

Your code has a small defect that works when you use < but not !=.
If you see str.charCodeAt(i+1); this code is checking one spot past the end of the string on the last iteration and will return a NaN result.
If I provide the string "abce" it will check if f is < NaN. I believe NaN can't be compared to f's value so it doesn't go into the if statement. So it will keep the missing letter d that was found in the previous iterations which is stored in your string variable.
However, if you provide the !=, then with the same scenario it knows f != NaN and goes into the if statement. This then overwrite the actual missing letter and fails your FCC test case because it is replacing the missing d with f in your string variable.
To fix your code, simply change the for loop to end one iteration before the length of the string.
for (i = 0; i != str.length-1; i++) {
}

This is my method without using .charCodeAt() function :)
function fearNotLetter(str) {
var ind;
var final = [];
var alf =['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'];
str = str.split('');
ind = alf.splice(alf.indexOf(str[0]),alf.indexOf(str[str.length-1]));
for(var i=0;i<ind.length;i++){
if(str.indexOf(ind[i]) == -1){
final.push(ind[i]);
}
}
if(final.length != 0){
return final.join('');
}
return;
}
fearNotLetter("bcef");

My solution:
function fearNoLetter(str){
var j= str.charCodeAt(0);
for(var i=str.charCodeAt(0); i<str.charCodeAt(str.length-1); i++){
j = str.charCodeAt(i - str.charCodeAt(0));
if (i != j){
return String.fromCharCode(i);
}
}
}

My solution:
function fearNotLetter(str) {
let y = 0;
for (let i = str.charCodeAt(0); i < str.charCodeAt(str.length - 1); i++) {
if (str.charCodeAt(y) != i) {
return String.fromCharCode(i);
}
y++;
}
return;
}
console.log(fearNotLetter("ace"));

function fearNotLetter(str) {
let alpha = "abcdefghijklmnopqrstuvwxyz";
let alphabet = []
for(let j = 0; j< alpha.length; j++){
alphabet.push(alpha[j])
}
if (alphabet.length == str.length){
let result = undefined;
return result
}else{
const start =alphabet.indexOf(str[0])
let end = (str.length)-1
const stop = alphabet.indexOf(str[end])
const finish = alphabet.slice(start,stop)
let result = finish.filter(item => !finish.includes(item) || !str.includes(item))
result = String(result)
return result
}
return result
}
console.log(fearNotLetter("abcdefghijklmnopqrstuvwxyz"));

String with the highest frequency of recurring letters in a word

This is a challenge for coderbyte I thought I'd try to do it using a different method for solving it than loops, objects. It passed but it isn't perfect. The directions for the challenge are:
Have the function LetterCountI(str) take the str parameter being passed and return the first word with the greatest number of repeated letters. For example: "Today, is the greatest day ever!" should return greatest because it has 2 e's (and 2 t's) and it comes before ever which also has 2 e's. If there are no words with repeating letters return -1. Words will be separated by spaces.
function LetterCountI(str){
var wordsAndLetters = {};
var count = 0;
var finalword;
str = str.split(" ");
for(var i = 0; i < str.length; i++){
wordsAndLetters[str[i]] = wordsAndLetters[str[i]] || 0;
}
function countWordLetters(strs){
strs = strs.split("");
var lettercount = {};
for(var i = 0; i <strs.length; i++){
lettercount[strs[i]] = lettercount[strs[i]] || 0;
lettercount[strs[i]]++;
}
return lettercount;
}
for(var words in wordsAndLetters){
wordsAndLetters[words] = countWordLetters(words);
var highestLetterFrequency = wordsAndLetters[words];
for(var values in highestLetterFrequency){
if(highestLetterFrequency[values] > count){
count = highestLetterFrequency[values];
finalword = words;
}
if(count !== 1){
return finalword;
}
}
}
return -1;
}
LetterCountI("today is the greatest day ever!");
Sorry if some of the variable names are confusing I've been up for far too long trying to figure out what I did wrong. If you use the parameters at the bottom of the code it returns 'greatest' like it should however change the parameters to
LetterCountI("toddday is the greatttttest day ever!");
and it logs 'toddday' when it should log 'greatttttest'. Is my code completely wrong? I realize if the parameters were ("caatt dooog") it should log 'caatt' since there are 4 recurring letters but I'm not worried about that I just am concerned about it finding the most recurrence of one letter(but by all means if you have a solution I would like to hear it!). Any changes to the variables if needed to make this code more readable would be appreciated!

The problem with your code is the positioning of the following section of code:
if(count !== 1){
return finalword;
}
Move it from where it currently is to just before the return -1, like so:
for(var words in wordsAndLetters){
wordsAndLetters[words] = countWordLetters(words);
var highestLetterFrequency = wordsAndLetters[words];
for(var values in highestLetterFrequency){
if(highestLetterFrequency[values] > count){
count = highestLetterFrequency[values];
finalword = words;
}
}
}
if(count !== 1){
return finalword;
}
return -1;
The problem with your original code is that your were returning the first word that had repeating characters, which meant your code didn't get far enough to check if any subsequent words had more repeating characters.
Also, just for fun, here is my alternative solution.

Here you go
Array.prototype.getUnique = function(){
var u = {}, a = [];
for(var i = 0, l = this.length; i < l; ++i){
if(u.hasOwnProperty(this[i])) {
continue;
}
a.push(this[i]);
u[this[i]] = 1;
}
return a;
}
function LetterCountI(str){
var temp = str.split(" ");
var final = '', weight = 0;
for(var i = 0; i < temp.length; ++i) {
var word = temp[i].split("");
if(word.getUnique().length < word.length) {
var diff = word.length - word.getUnique().length;
if(diff > weight){
weight = diff;
final = temp[i];
}
}
}
return final;
}
console.log(LetterCountI("Catt dooog"));
console.log(LetterCountI("toddday is the greatttttest day ever!"));

Viva LinQ !!!!!
var resultPerWord = new Dictionary<string, int>();
var S = "toddday is the greatttttest day ever!";
foreach(var s in S.Split(' '))
{
var theArray =
from w in s
group w by w into g
orderby g.Count() descending
select new { Letter = g.Key, Occurrence = g.Count() };
resultPerWord.Add(s, theArray.First().Occurrence);
}
var r = "-1";
if (resultPerWord.Any(x => x.Value >1))
{
r = resultPerWord.OrderByDescending(x => x.Value).First().Key;
}

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)

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.

Looping through arrays to check for patterns

I'm trying to loop through an array to check for a specific pattern but keep getting no output afterwards. Not sure what I've done wrong! I would appreciate any help!
I am testing for the pattern at or hat.
sample = ["cat fat hat mat", "that the who"]
searchTerm = prompt("Testing?");
function count(sample, searchTerm)
{
for (i=0;i<sample.length;i++)
{
if (sample[i].indexOf(searchTerm) == -1)
{
return 0;
}
return count(sample.substring(sample.indexOf(searchTerm) + searchTerm.length), searchTerm) + 1;
}
}
alert(count(sample, searchTerm));
Rehashed code
search = ["cat fat hat mat", "that the who"];
var pattern = prompt('Search?');
function count(sample, searchTerm)
{
var count, i;
count = 0;
for (i=0; i < sample.length; i++)
{
if (sample[i].indexOf(searchTerm) !== -1)
{
count++;
}
}
return count;
}
count(search, pattern);
I've redone everything and it still gives no output.

There are a couple of problems with this code. The most immediate one is you are calling substring on an array and not a string.
return count(sample.substring ...
Likely you meant to say
return count(sample[i].substring ...
The second issue though is that you need to divide the logic up a bit. You need to divide it up into sections that count the occurrences in a word and that which iterates through the array. Today they are intertwined and results in odd behavior because you end up passing non-arrays to places expecting arrays
function count(sample, searchTerm) {
var num = 0;
for (i=0;i<sample.length;i++) {
var current = sample[i];
var index = current.indexOf(searchTerm);
while (index >= 0) {
num++;
index = current.indexOf(searchTerm, index + 1);
}
}
return num;
}
Working Fiddle: http://jsfiddle.net/wrNbL/

You don't need to use recursion here, just iterate through the array once counting if the search term matches.
function count(sample, searchTerm)
{
var count, i;
count = 0;
for (i=0; i < sample.length; i++)
{
if (sample[i].indexOf(searchTerm) !== -1)
{
count++;
}
}
return count;
}