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semi-space copy allocation failed, javascript heap out of memory

I was wondering why the below code returns a memory allocation error?
var countValidWords = function(sentence) {
let words = [];
for(let i = 0; i < sentence.length; ++i){
let word = '';
while(sentence[i] !== ' '){
word += sentence[i];
++i;
}
if(word !== '')
words.push(word);
}
console.log(words);
};
I'm simply trying to build an array of words from the inputted sentence (words can be separated by more than one space).

If the sentence doesn't end with a space, the while loop never ends, because it doesn't check if it has gone past the end of the string. As a result, you go into an infinite loop appending undefined to word, until you run out of memory.
Add a check that i is within the string length there.
var countValidWords = function(sentence) {
let words = [];
for(let i = 0; i < sentence.length; ++i){
let word = '';
while(i < sentence.length && sentence[i] !== ' '){
word += sentence[i];
++i;
}
if(word !== '')
words.push(word);
}
console.log(words);
};

The while clause needs to check for i getting out of bounds:
while(i < sentence.length && sentence[i] !== ' ')

How about this? It's not the most optimized version you might find but it works.
You had an infinite loop on the while.
var countValidWords = function(sentence) {
let words = [];
let word = ''
for(let i = 0; i < sentence.length; ++i){
if(sentence[i] != '' && sentence[i] != ' '){
word += sentence[i];
}
if(sentence[i] == ' ' || sentence[i + 1] === undefined){
words.push(word);
word = '';
}
}
console.log(words);
};

Changing the value of an index inside an array

I'm stuck with this problem where I need to eliminate a letter from each of the array values and change the first letter if it is in index 0. My current approach is this:
function capital(arr, char){
let str = "";
let result = "";
for (let i = 0; i < arr.length; i++){
str = arr[i] + (i < arr.length - 1 ? ",": "");;
for (let j = 0; j < str.length; j++){
if (str[j] === char){
result += "";
if (str[j] === char){
result += (j === 0? "A": "");
}
else {
result += str[j];
}
}
}
console.log(result);
}
capital(["doritos","sandking","bandana", "demand"], "d");
The program should eliminate each occurrence of letter d found in the strings and change the index 0 to letter A if the d is in index 0.
The current result is
Aoritos,sanking,banana,Aeman
but it should look like
Aritos,sanking,banana,Aman
The requirement does not allow any use of built in functions, and the actual program requires the letters to be case insensitive, but I can work on it by substituting codes and adding couple if elses, I just need help with the changing the index 0 condition. Any help would be appreciated, thank you!

You can check if your input char is at first index str.indexOf(char) then simply add prefix "A" to your string without first char i.e str.substring(1)
function capital(arr, char) {
let str = "";
let result = "";
for (let i = 0; i < arr.length; i++) {
str = arr[i];
if(str.indexOf(char) === 0) {
result = 'A' + str.substring(1);
}else {
result = str;
}
console.log(result);
}
}
capital(["doritos", "sandking", "bandana", "demand"], "d");

this logic may help you
function capital(arr, char) {
return arr.map(e => {
let isFirst = false;
if (e[0] == "d") isFirst = true;
e = e.replace(new RegExp(char, 'ig'), '');
if (isFirst)
e = e.replace(e.charAt(0), "A");
return e;
});
}
console.log(capital(["doritos", "sandking", "bandana", "demand"], 'd'))

You can iterate through each letter of your word and update the first letter if it is matched with passed letter and at 0 index, for other index if it doesn't match with passed letter add to your result.
const capitalize = ch => {
const letter = ch.charCodeAt(0);
if(letter >= 97 && letter <= 122) {
return String.fromCharCode(letter - 32);
}
return ch;
}
const capital = (words, ch) => {
let result = '';
for(let i = 0; i < words.length; i++) {
let newWord = '';
for(let j = 0; j < words[i].length; j++) {
if(capitalize(words[i][j]) === capitalize(ch) && j === 0) {
newWord = 'A';
}
if(capitalize(words[i][j]) !== capitalize(ch)) {
newWord += words[i][j];
}
}
result += newWord;
if(i < words.length - 1 ) {
result += ',';
}
}
return result;
}
const result = capital(["doritos","sandking","bandana", "demand", "sandDking"], "d");
console.log(result);

how to record index after loop in java script

im about to make a program that the output obtain the domain of the email. how to do that without built-in function such as .map .filter .reduce .split .join .indexOf .findIndex .substring
i do search and many answer said i have to use for loop, i can find the "#" and find the "." but how to make those string between "#" and "." to be an output
ex: input = mybest#gmail.com
output = gmail
input = triple2#yahoo.com
output = yahoo
let input = "mybest#gmail.com"
let output = ""
let begin = ""
let end = ""
for (let i = 12; i<input.length; i++){
if(input[i] == "#"){
begin += input[i+1]
}
}
for (let j = 0; j<input.length; j++){
if(input[j] == "."){
end += input[j-1]
}
}

function tmp() {
let input = "mybest#gmail.com"
let output = []
let didReachAt = false
for (let i = 0; i < input.length; i++) {
if(input[i] == "#") {
didReachAt = true
} else if(input[i] == '.'){
break
} else if(didReachAt) {
output.push(input[i])
}
}
return output.join('')
}
console.log(tmp())
The key here is to only start appending to the output when you come across an # and stop appending when you come across a . When you come across a ., you know you have your output so you can break and return the value.

let input = "mybest#gmail.com"
for (let i = 0;i<input.length;i++) {
if (input[i] === '#') {
var start = i+1
} else if (input[i] === '.') {
var end = i+1
return
}
}
let output = input.substring(start,end)
Simple solution although this does not account for errors such as a false input

So i can see that you started with javascript, so you can learn better using code like this.
var input = "triple2#yahoo.com"
var domainWithoutCom = input.split('#')[1].split('.')[0]
console.log(domainWithoutCom)
split can do the job for you!
EDIT:
Sorry about that, so i will create one version to you check my answer back :)
Check this, i think that can be improve, but can give you an idea :)
var email = 'calvin.nunes#yahoo.com';
var count = 0;
var domain = '';
var findFirst = false;
var searching = true;
do {
var char = email[count];
if (findFirst) {
if (char == '.') {
searching = false;
} else {
domain += char;
}
}
else {
if (char == '#') {
findFirst = true;
}
}
count += 1;
} while (searching == true);
console.log(domain)

Longest substring in alphabetical order Javascript

Seeing all the people talking about longest substring in alphabetical order in Python, I have decided to try it in JS.
The function should look for the longest substring inside a given string, where letters are ordered alphabetically.
Here is what I have:
var s = 'azcbobobegghakl'
function substringChecker(s) {
var longestSub = "";
for (var i = 0; i < s.length; i++) {
var count = 0;
var currSub = "";
while((i+count)<=s.length){
var curr = i+count;
var next = curr+1;
var prev = curr-1;
if(curr !== s.length-1) {
if(s[curr] <= s[next]){
currSub += s[curr]
} else {
break;
}
} else {
if(s[curr]>s[prev]) {
currSub += s[curr];
}
}
count++;
}
if(currSub.length >= longestSub.length) {
longestSub = currSub;
}
};
return longestSub;
}
var result = substringChecker(s);;
console.log(result);
The funny thing it works great for all test cases I can come up with, but this one. The result should be "beggh" but it is "begg" instead. Why is the h not showing up, what am I missing?

The algorithm can be linear, I think you are overcomplicating it placing loops inside loops.
I would use something like
function substringChecker(s) {
var longestSub = "",
length = 0,
start = 0,
prev = s[0];
for (var i = 1; i <= s.length; ++i) {
if(i == s.length || s[i] < prev) {
if(length < i-start) {
longestSub = s.substring(start, i);
length = i-start;
}
start = i;
}
prev = s[i];
};
return longestSub;
}
document.write(substringChecker('azcbobobegghakl'));

first I made list of A-z
then check each letter and compare it with the next letter and save it in subString and...
function longest(str) {
//handle the case str is just one letter
if (str.length === 1) return str;
// create a list of alphabet A to Z
const alphabets = [...Array(26)].map(_ => String.fromCharCode(i++), (i = 97));
let longString = "";
let subSting = "";
for (let x = 0; x < str.length; x++) {
let char = str.charAt(x);
const nextChar = str.charAt(x + 1);
let charIndex = alphabets.findIndex(alphabet => alphabet === char);
let nextCharIndex = alphabets.findIndex(alphabet => alphabet === nextChar);
if (nextCharIndex >= charIndex) {
subSting = subSting + nextChar;
} else {
if (!subSting.length) {
subSting = subSting + char;
}
longString = subSting.length > longString.length ? subSting : longString;
subSting = "";
}
}
return longString;
}
console.log(longest("zyba"));

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.