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Javascript - Find opening and closing bracket positions on a string?

I'm making a calculator for a site project of mine where you can type your entire expression before resolving, for example: 2+3*4 would return 14, 22-4 would return 18, 20+5! would return 140, and so on.
And that works for simple expressions like the ones I showed, but when I add brackets the code breaks.
So a simple expression like (2+3)! that should return 120 actually returns 10 or 2+3!.
my original ideia to make even the basic 2+3! work was to separate the string in math simbols and the rest. so it would separate in this case it would separate it into 2, + and 3!; where it would find the symbol and resolve just that part. And that's why it solves 10 instead of not working.
But after trying to solve I couldn't make the code work except in a extremely specific situation, so I decided to redo the code and post this here in case someone could help me out.
This is the function that I'm currently using to prepare my string for evaluation:
function sepOperFat(){
//2+3! it's working
//1+(2-(2+2)+3)! want that to work in the end
var value = document.calculator.ans.value;
var operandoPos = ['0'];
var operandoInPos = [''];
var paraResolver = [];
for(i = 0; i <= value.length; i++){
//check if value[i] is equal to +, -, ×, ÷, * & /
if(value[i] == '+' || value[i] == '-' || value[i] == '×' || value[i] == '÷' || value[i] == '*' || value[i] == '/'){
operandoPos.push(i);
operandoInPos.push(value[i]);
}
}
paraResolver.push(value.slice(operandoPos[0], operandoPos[1]));
for(var total = 1; total <= operandoPos.length; total++){
paraResolver.push(value.slice(operandoPos[total] + 1, operandoPos[total + 1]));
}
document.calculator.ans.value = '';
for(var total = 0; total <= paraResolver.length - 2; total++){
if(paraResolver[total].includes('!')){
document.calculator.ans.value += "factorial(" + paraResolver[total] + ")";
}else{
document.calculator.ans.value += paraResolver[total];
}
document.calculator.ans.value += operandoInPos[total + 1];
}
}
document.calculator.ans.value is the name of the string where i have the expression.
operandoPos is the position on the string where a symbol is at.
operandoInPos is the symbol (I maybe could have used value.charAt(operandoPos) for that too).
paraResolver is the number that I will be solving (like 3).
factorial( is the name of my function responsible for making the number factorial.
the function doesn't have a return because I still want to solve inside the document.calculator.ans.value.
to resolve the equation I'm using document.calculator.ans.value = Function('"use strict"; return '+ document.calculator.ans.value)(); that activates when I press a button.
And yeah, that's it. I just want a function capable of knowing the difference between (2+3)! and 2+(3)! so it can return factorial(2+3) instead of (2+factorial(3)).
Thank you for your help.

Your biggest problem is going to be that order of operations says parentheses need to be evaluated first. This might mean your code has to change considerably to support whatever comes out of your parentheses parsing.
I don't think you want all of that handled for you, but an approach you can take to sorting out the parenthesis part is something like this:
function parseParentheses(input) {
let openParenCount = 0;
let myOpenParenIndex = 0;
let myEndParenIndex = 0;
const result = [];
for (let i = 0; i < input.length; i++) {
if (input[i] === '(') {
if (openParenCount === 0) {
myOpenParenIndex=i;
// checking if anything exists before this set of parentheses
if (i !== myEndParenIndex) {
result.push(input.substring(myEndParenIndex, i));
}
}
openParenCount++;
}
if (input[i] === ')') {
openParenCount--;
if (openParenCount === 0) {
myEndParenIndex=i+1;
// recurse the contents of the parentheses to search for nested ones
result.push(parseParentheses(input.substring(myOpenParenIndex+1, i)));
}
}
}
// capture anything after the last parentheses
if (input.length > myEndParenIndex) {
result.push(input.substring(myEndParenIndex, input.length));
}
return result;
}
// tests
console.log(JSON.stringify(parseParentheses('1!+20'))) // ["1!+20"]
console.log(JSON.stringify(parseParentheses('1-(2+2)!'))) // ["1-",["2+2"],"!"]
console.log(JSON.stringify(parseParentheses('(1-3)*(2+5)'))) // [["1-3"],"*",["2+5"]]
console.log(JSON.stringify(parseParentheses('1+(2-(3+4))'))) // ["1+",["2-",["3+4"]]]
this will wrap your input in an array, and essentially group anything wrapped in brackets into nested arrays.
I can further explain what's happening here, but you're not likely to want this specific code so much as the general idea of how you might approach unwrapping parenthesis.
It's worth noting, the code I've provided is barely functional and has no error handling, and will behave poorly if something like 1 - (2 + 3 or 1 - )2+3( is provided.

Freecodecamp - Counting Cards

I know there are a number of ways to complete this challenge and I can simply a different approach to pass the requirement however I am struggling to understand what's wrong with my code.
Challenge - https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/basic-javascript/counting-cards/
Any help would be much appreciated.
var count = 0;
function cc(card) {
// Only change code below this line
if (card < 7){
count++;
} else if (card < 10){
count+=0;
} else (count--;)
if (count > 0){
return count + " Bet";
} else (
return count + " Hold";
)
// Only change code above this line
}
// Add/remove calls to test your function.
// Note: Only the last will display
cc(2); cc(3); cc(7); cc('K'); cc('A');

Firstly, when you try to execute your code, you should be seeing a Syntax Error, pointing to the semicolon in (count--;). The reason for this is: else expects a statements, if it sees a parenthesis it means the statement is an expression, and in an expression, semicolons can't appear inside parentheses (this is rather simplified). The correct way to write it is either without parentheses (generally frowned upon) as else count--;, or with curly braces: else { count--; }.
When you fix that error, there will be another one of the same kind, as you seem to systematically use parentheses instead of curly braces after else.
After that, your code kind of works. There's questionable comparisons of card, that can be a letter or a number, with an integer, but it coincidentally works the way you hope it does (because 'K', 'Q' and 'J' happen to be evaluated as greater than 7 and 10.) It would be better to not rely on such magic, and have a translation table between letters and values - or at least, if you're going to rely on magic, comment so that readers are aware you are aware of the magic. Also, count+=0 is a void statement, it does nothing, and could have been left out. That leaves you with an empty else if, but that's not an error. However, it would probably be much more readable if you had if (card < 7) { count--; } else if (card >= 10) { count++; }.

stupid mistake's.
answer is below:
var count = 0;
function cc(card) {
// Only change code below this line
if (card < 7){
count++;
} else if (card < 10){
count;
} else {count--;
}
if (count > 0){
return count + " Bet"
} else {
return count + " Hold"
}
// Only change code above this line
}

Write a JavaScript function to find longest substring in a given a string without repeating characters

"Write a JavaScript function to find longest substring in a given a string without repeating characters."
Here's what I tried, but it doesn't print anything
function sort(names) {
let string = "";
let namestring = names.split("");
for(let i = 0; i < namestring.length; i++) {
for(let j = 0; j < string.length; j++) {
if(string[j] != namestring[i]) {
string = string + namestring[i];
}
}
}
return string;
}
console.log(sort("google.com"));
What's wrong?

function sort(names)
{
string="";
ss="";
namestring=names.split("");
for(j=0;j<namestring.length;j++) {
for(i=j;i<namestring.length;i++) {
if(string.includes(namestring[i]))
break;
else
string+=namestring[i];
}
if(ss.length<string.length)
ss=string;
string="";
}
return ss;
}
console.log(sort("google.com"));
It's o(n^2) complexity but try this(may be o(n^3) if contains function take o(n) complexity)
function sort(names)
{
string="";
ss="";
namestring=names.split("");
for(j=0;j<namestring.length;j++) {
for(i=j;i<namestring.length;i++) {
if(string.includes(namestring[i])) // if contains not work then
break; //use includes like in snippet
else
string+=namestring[i];
}
if(ss.length<string.length)
ss=string;
string="";
}
return ss;
}
console.log(sort("google.com"));

What are you expecting the answer to be here? Should it be "ogle.com" or "gle.com"? If the first, the below should get you there, if the latter, update the tested = name.charAt(i) in the else to tested = "".
So a few things to note, though you're more than welcome to do as you wish:
1) the function name. This isn't doing a "sort" as far as I can tell, so if this is for your use (or any reuse. Basically, anything more than a one off homework assignment), you may want to rename it to something you'd actually remember (even the example I give is probably not completely best as "pick longest substring" is non-descriptive criteria).
2) variable naming. string and namestring may mean something to you here, but considering we're trying to find the longest substring (with the no double characters) in a string, I felt it was better to have the one we're checking against (tested) and the one we're storing to return later (longest). It helps make sense as you're reading through the code as you know when you are done with a checked string (tested), you want to compare if it is greater than the current longest substring (longest) and if it is bigger, you want it to be the new longest. This will save you a ton of headache to name variables to things that'll help when designing your function as you can get it as close to requirements written down as possible without trying to do some form of substitution or worse, forgetting which variable holds what.
I don't know what you want the result to be in the event that tested length is the same as longest length. Currently I have it set to retain, if you want the most recent, update the check to >=.
Beyond that, I just iterate over the string, setting to the currently tested string. Once double characters are met, I then see if what I just generated (tested) is larger than the current longest and if it is, it is now the longest. Once I finish looping across the string, I have to do the current vs longest check/set again as otherwise, it'd make the final tested meaningless (it went outside the loop before another double character situation was hit).
function pickLongestSubstring(name) {
let tested = "";
let longest = "";
for (let i = 0; i < name.length; i++) {
if (tested.length == 0 || tested.charAt(tested.length - 1) != name.charAt(i)) {
tested += name.charAt(i);
}
else {
if (tested.length > longest.length) {
longest = tested;
tested = "";
}
}
}
if (tested.length > longest.length) {
longest = tested;
}
return longest;
}
console.log(pickLongestSubstring("google.com"))
console.log(pickLongestSubstring("example.com"))

This is a recursive loop that should get the longest string. Uses sort to determine longest string. Works, even if multiple instances of same repeat char.
function longestWithoutRepeat(testString, returnString){
var returnString = returnString || "";
for(var i = 0; i < testString.length; i++) {
if(i > 0){
if(testString[i] == testString[i-1]) {
var testStringArray = testString.split(testString[i] + testString[i-1]);
testStringArray.sort(function(firstString, nextString){ return nextString.length - firstString.length})
returnString = testStringArray[0];
longestWithoutRepeat(testStringArray[0], returnString);
}
} else {
returnString = testString
}
}
return returnString;
}
console.log(longestWithoutRepeat("oolong"));
console.log(longestWithoutRepeat("google.com"));
console.log(longestWithoutRepeat("diddlyougotoofarout"));

Recursion and Loops - Maximum Call Stack Exceeded

I'm trying to build a function that adds up all the numbers within a string... for example, 'dlsjf3diw62' would end up being 65.
I tried to be clever and put together a recursive function:
function NumberAddition(str) {
var numbers='1234567890';
var check=[];
str=str.split[''];
function recursive(str,check) {
if (str.length==0)
return check;
else if (numbers.indexOf(str[0])>=0)
{
for (i=0;i<str.length;i++){
if (numbers.indexOf(str[i])<0)
check.push(str.slice(0,i));
str=str.slice(i);
return recursive(str,check);
}
}
else
str.shift();
return recursive(str,check);
}
You'll see that I'm trying to get my numbers returned as an array in the array named check. Unfortunately, I have a maximum call stack size exceeded, and I'm not sure why! The recursion does have a base case!! It ends once str no longer has any contents. Why wouldn't this work? Is there something I'm missing?
-Will

You can achieve the same thing with a far easier solution, using regular expressions, as follows:
var str = 'dlsjf3diw62';
var check = str.match(/\d+/g); // this pattern matches all instances of 1 or more digits
Then, to sum the numbers, you can do this:
var checkSum = 0;
for (var i = 0; i < check.length; i++) {
checkSum += parseInt(check[i]);
}
Or, slightly more compact:
var checkSum = check.reduce(function(sum, num){ return sum + parseInt(num) }, 0);

The reason your recursion doesn't work is the case where you do enter the for loop, because you've found a digit, but the digits continue to the end of the string. If that happens, the return inside the for loop never happens, and the loop ends. After that, the .shift() does not happen, because it's in that else branch, so you return re-process the same string.
You shouldn't solve this particular problem that way, but the code makes a good example of the anti-pattern of having return statements inside if bodies followed by else. Your code would be clearer (and would work) if it looked like this:
function recursive(str, check) {
if (str.length == 0)
return check;
if (numbers.indexOf(str[0]) >= 0) {
// Find the end of the string of digits, or
// the end of the whole thing
for (var i = 0; i < str.length && numbers.indexOf(str[i]) >= 0; i++);
check.push(str.slice(0, i));
str = str.slice(i);
return recursive(str, check);
}
// A non-digit character
str.shift();
return recursive(str, check);
}
In that version, there are no else clauses, because the two if clauses always involve a return. The for loop is changed to simply find the right value of "i" for the subsequent slicing.
edit — one thing this doesn't fix is the fact that you're pushing arrays into your "check" list. That is, the substring "62" would be pushed as the array ["6", "2"]. That's not a huge problem; it's solved with the addition of a .join() in the right place.

Codecademy lesson troubles

I am learning JavaScript through Codecademy, but I have an issue. The code below is supposed to search through the text variable for my name in the myName variable and then push all of the individual letters to the hits array. The code that I have written is not correct but Codecademy says that it is correct and is going to let me move on in the lesson.
I have been trying to solve the issue that I am having with no luck. The problem is that when I run the hits.push(text); line it will output the entire variable but I have tried hits.push(text[i]); and get undefined for the result. Can someone please help me understand where I have made the mistake?
/*jshint multistr:true */
var text = "XsddfasASSFABrandonSFsdfdasBrandonsddfadfaBrandon";
var myName = "Brandon";
var hits = [];
for (i=0; i<=text.length;i++){
if (text[i]===myName[i]){
for(var x=i; x<i+myName.length;x++){
hits.push(text);
}
}
}
if (hits.length===0){
console.log("Your name wasn't found!");
} else {
console.log(hits);
}

The best way I can think to explain your mistake is simply by walking through a bit of the logic of what you have written.
for (i=0; i<=text.length;i++){
Your for loop will iterate i for as many characters as there are in your text variable, so: 49 times.
if (text[i]===myName[i]){
The first run through your for loop, where i=0, you are checking to see if text[0] is strictly equal to myName[0]. text[0] = X and myName[0] = B. The strictly equals condition is not met, so the loop proceeds to increment i repeat: text[1] = s and myName[1] = r. This continues 47 more times, and the condition is never met. myName[i] is undefined after the first 7 loops.

Normally you would do this kind of thing using indexOf, match, search, substr or substring, which are all string methods.
However for the purpose of this exercise you can do:
var text = "XsddfasASSFABrandonSFsdfdasBrandonsddfadfaBrandon";
var myName = "Brandon";
var hits = [],
namePosition = 0;
for (var i = 0; i < text.length; i++) {
if (text[i] === myName[namePosition]) {
hits.push(text[i]);
namePosition ++;
if (hits.length === myName.length) {
break;
}
}
else {
namePosition = 0;
hits = [];
}
}
if (hits.length === 0) {
console.log("Your name wasn't found!");
} else {
console.log(hits);
}​
(See it working at http://jsfiddle.net/wCWxr/1/). The problems with your original code include:
you try to compare text[i] to myName[i] but the indices of the two strings won't match up.
you try to push the entire string text into hits instead of one character at a time
your logic doesn't deal with the possibility that the beginning but not the end of myName is in text, e.g. if text was aerwerBrasdfsgars
My suggestion fixes this by recording (with namePosition) what position we are currently at within the string myName, and incrementing that when we find a character in text that matches the relevant character in myName. If the characters do not match then it's not a true hit, so we reset hits = [] and namePosition = 0. If the characters all match then hits eventually reaches the length of myName and so we break out of the loop.

If you are trying to find if myName is in text here is what you do:
RegExp:
var pattern = new RegExp(myName);
if (pattern.test(text)){
console.log(myName);
}else {
console.log("Your name wasn't found!");
}
indexOf:
if (text.indexOf(myName) != -1){
console.log(myName);
}else {
console.log("Your name wasn't found!");
}

if (text[i]===myName[i]){
this line should create an error, because myName[i] is not the first letter of myName.
if (text[i]===myName[0]){
Change to this line should work.