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Javascript recursive example

I recently had an interview where you had to recursively go over a string, and if it contained an AB || BA || CD || DC, it had to be deleted from the array. You would recursively go over this as deleting the CD from ACDBB would give you an AB which you would then have to delete to return a B as a string.
This is what I have, and when I test it out, I see it comes up with the right answer deep in the loops, but it never populates back to the top.
What am I missing?
const LETTERS = [/AB/g, /BA/g, /CD/g, /DC/g];
const stringGame = (string) => {
let newString = '';
if(string.length <= 1) return string;
LETTERS.forEach(regExToCheck => {
if(string.match(regExToCheck)) {
newString = string.replace(regExToCheck, '')
}
stringGame(newString);
})
return newString
}
// Expect answer: CAACC
console.log(stringGame('ABDCABCABAAABCCCD'))

Move the recursion,
return the recursion, and
Add an essential condition to end recursion:
const LETTERS = [/AB/g, /BA/g, /CD/g, /DC/g];
const stringGame = (string) => {
let newString = '';
if (string.length <= 1) return string;
LETTERS.forEach(regExToCheck => {
if (string.match(regExToCheck)) {
newString = string.replace(regExToCheck, '')
}
})
// Moved, returned, and ended recursion
return ('' === newString) ? string : stringGame(newString);
}
// Expect answer: CAACC
console.log(stringGame('ABDCABCABAAABCCCD'))
When recursing the function, i.e., when executing the function within itself, the return'ed value of that execution needs to be used, typically return'ed:
return stringGame(newString);
...in order for it to bubble back up the levels of recursion.
Also, since the replace() function is using a regex global replacement flag (g) all, for example, AB's, are being replaced in a single execution, so there's no need to recurse within the forEach() loop:
LETTERS.forEach(regExToCheck => {
if(string.match(regExToCheck)) {
newString = string.replace(regExToCheck, '')
}
stringGame(newString); // <-- double oops
})
Rather, recurse after the loop, AND return the value of the execution:
LETTERS.forEach(regExToCheck => {
if(string.match(regExToCheck)) {
newString = string.replace(regExToCheck, '')
}
})
return stringGame(newString);
One more principle of recursive functions is providing exit strategies—define what conditions to end the recursion, bubble back up, and return a final result.
The only exit condition provided is:
if (string.length <= 1) return string;
Surely if there's only one character left no match will be found—a proper time to end recursion. However, and more importantly, there also needs to be a path to exit when there are more than one characters but no matches can be found.
There are multiple ways to determine this but in the case provided in the question the most expedient is when after looping through the regex/replace routine no changes were made to the string. Since with every recursion the stringGame() function instantiates an empty string (newString) if, after looping through the regex/replace routine, newString is still an empty string, this indicates there were no matches found and it's time to end recursion:
if ( '' === newString ) {
return string;
}
else {
return stringGame(newString);
}
...otherwise, recurse.

The stringGame function doesn't have side effects, so the line in the loop here:
stringGame(newString);
doesn't do anything - you need to communicate the result of the recursive call back to the outer level.
A nicer way to approach this would be to combine the LETTERS into a single regular expression, and then to replace all matches with the empty string until the replacement produces no changes.
const LETTERS = [/AB/g, /BA/g, /CD/g, /DC/g];
const pattern = new RegExp(
LETTERS
.map(re => {
const str = String(re);
return str.slice(1, str.length - 2);
})
.join('|'),
);
const stringGame = (string) => {
const newString = string.replace(pattern, '');
return newString === string
? string
: stringGame(newString);
}
// Expect answer: CAACC
console.log(stringGame('ABDCABCABAAABCCCD'))
(If you want to use the global flag in the constructed pattern, you'll have to account for the .lastIndex)

Check if a string is made up of multiple occurences of a substring

(JavaScript)
So, I need a function checkString(str, substr) that checks whether a string is made out of multiple occurences of a given substring.
Examples:
checkString("abc", "abc") -> true
checkString("abcabcabc", "abc") -> true
checkString("abcdef", "abc") -> false
checkString("abcab", "abc) -> true
Could someone help me out?

If whitespaces don't matter, this is a solution:
const checkString = (bigString, subString) => {
const split = bigString.split(subString);
const onlyOccurances = split.filter(v => v === '');
return split.length === onlyOccurances.length
}
checkString("abc", "abc") // true
checkString("abcabcabc", "abc") // true
checkString("abcdef", "abc") // false
bigString.split(subString) will split the big string into an array of empty strings if there's perfect match, and the length of it will be exactly how many occurances there are. If any of the values in the array are not empty strings, it means there is not a perfect match so there will be a difference between the length of the filtered by empty and the length of the splited values.
Hope it makes sense.

This method will return an object and found will be true if the value is found in the string and multipleFound will be true if found more than once;
const checkString = function(str, v) {
let found = false,
multi = false,
index;
index = str.indexOf(v);
if (index !== -1) {
found = true;
index = str.indexOf(v, index + v.length);
if (index !== -1) {
multi = true;
}
}
return {
found : found,
multipleFound : multi
};
};

This would be one way to check against the pattern abc:
const rx=/^(abc)+$/;
console.log(["abc","abcabcabc","abcdef"].map(t=>
`${t} ${rx.test(t)}`))

My code is not comparing two different strings correctly

I am attempting to solve this codewars problem:
Complete the function scramble(str1, str2) that returns true if a
portion of str1 characters can be rearranged to match str2, otherwise
returns false.
examples:
scramble('rkqodlw', 'world') ==> True
scramble('cedewaraaossoqqyt', 'codewars') ==> True
scramble('katas', 'steak') ==> False
This is my attempt:
function scramble(str1, str2) {
let obj1 = {};
let obj2 = {};
for (el of str1) {
obj1[el] = (obj1[el] || 0) + 1;
}
for (el of str2) {
obj2[el] = (obj2[el] || 0) + 1;
}
for (el in obj2) {
if (!(el in obj1)) return false;
}
return true;
}
I am converting the strings to objects, and then comparing the keys to each other. My code passes about 90% of all the tests on Codewars, but then it does not pass the other 10% and they do not show what the test inputs are unfortunately.
My hunch is that there are a few edge cases that this code is not catching. Any help would be appreciated.

You will need to handle the case when str2 has more instances of a letter than str1.
For example:
scramble("a", "aa")
Which should evaluate to false (not enough 'a's in "a" to form "aa").

You need to handle the amount of characters. Now you just checking whether the character in str2 also exists in str1.
So instead of:
for (el in obj2) {
if (!(el in obj1)) return false;
}
Try:
for (let [key, value] of Object.entries(obj2)) {
if (obj1[key] === undefined || obj1[key] < value) return false;
}
Which means that if obj1[key] doesn't exist or it has less occurrences than obj2[key] it will return false.

this code maybe helps
function scramble(str1, str2) {
let occurences = str1.split("").reduce((arr, cur) => { arr[cur] ? arr[cur]++ : arr[cur] = 1; return arr; }, {});
console.log(occurences);
return str2.split("").every((character) => --occurences[character] >= 0);
}
console.log(scramble("awpoirwled", "world"));

The algorithm I would use is to take each letter of the match string, and if its got an equivalent in the searched string, remove it from there and continue, otherwise abort if there is no match.
If you reach the end of the match string without aborting then you have a match

eval is not calculate if zero before the number

var values = ["50.00024+40.04005+0.1", "0050.00024+040.04005+0.1"];
for (var i=0; i<values.length; i++) {
if(values[i].indexOf('+')>0 || values[i].indexOf('-')>0 || values[i].indexOf('*')>0 || values[i].indexOf('/')>0){
try{
var evaluated = eval(values[i]);
if(typeof evaluated === 'number'){
console.log(evaluated);
}
}catch (e){
console.error(e)
}
}
}
I have some math actions, it's could be plus, minus or other actions, and I need to take result for this actions. I use eval for this. But if I have zero before number like 005,75 eval is not working. How can I calculate this?

You can split the strings and parse the numbers, and then make them into a string again to use eval
var values = ["50.00024+40.04005+0.1", "0050.00024+040.04005+0.1"];
values.forEach(function(value){
var newValue = value.split(/([\+\-\*\/])/).map(a => parseFloat(a) || a).join('');
var evaluated = eval(newValue);
console.log(value,"==", evaluated);
});

There are various libraries like math.js that can be used to evaluate expressions:
console.log(math.eval("0050.00024+040.04005+0.1"))
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/3.16.5/math.min.js"></script>

you should convert it to float (ex. parseFloat("005.75") = 5.75 instead of evaluating strings

Given code at Question you can use .match() to get number values, .map() and .reduce() to add the values
var values = ["50.00024+40.04005+0.1", "0050.00024+040.04005+0.1"];
var res = values.map(function(n) {
return n.match(/[\d.]+/g).reduce(function(a, b) {
return a + Number(b)
}, 0)
});
console.log(res);

How to check if a string only contains chars from an array?

I try to check if a word (wordToCheck) only consists of letters from an array (letters) and also contains every letter in the array only as often (or rather not more times than they are in the array) as it actually is inside of the array.
Here are examples of what the desired function should return:
checkIfWordContainsLetters("google", ["a","o","o","g","g","l","e","x"]) === true
checkIfWordContainsLetters("google", ["a","o","g","g","l","e","x"]) === false
How can I make this code work?
function checkIfWordContainsLetters(wordToCheck, letters) {
var lettersToString = letters.toString();
var lettersTrimmed = lettersToString.replace(/,/gi, "?");
var regEx = new RegExp(lettersTrimmed, "gi");
if (wordToCheck.match(regEx)!== null) {
return true;
}
else return false;
}

You could use this ES6 function:
function checkIfWordContainsLetters(wordToCheck, letters){
return !letters.reduce((a, b) => a.replace(b,''), wordToCheck.toLowerCase()).length;
}
console.log(checkIfWordContainsLetters("google", ["a","o","o","g","g","l","e","x"]));
console.log(checkIfWordContainsLetters("google", ["a","o","g","g","l","e","x"]));
The idea is to go through each letter in the letters array, and remove one (not more!) occurrence of it in the given wordToCheck argument (well, not exactly in it, but taking a copy that lacks that one character). If after making these removals there are still characters left over, the return value is false -- true otherwise.
Of course, if you use Internet Explorer, you won't have the necessary ES6 support. This is the ES5-compatible code:
function checkIfWordContainsLetters(wordToCheck, letters){
return !letters.reduce(function (a, b) {
return a.replace(b, '');
}, wordToCheck.toLowerCase()).length;
}
console.log(checkIfWordContainsLetters("google", ["a","o","o","g","g","l","e","x"]));
console.log(checkIfWordContainsLetters("google", ["a","o","g","g","l","e","x"]));

As long as it is not the best solution for long strings for which using some clever regex is definitely better, it works for short ones without whitespaces.
function checkIfWordContainsLetters(word, letters){
word = word.toLowerCase().split('');
for(var i = 0; i < letters.length; i++) {
var index = word.indexOf( letters[i].toLowerCase() );
if( index !== -1 ) {
// if word contains that letter, remove it
word.splice( index , 1 );
// if words length is 0, return true
if( !word.length ) return true;
}
}
return false;
}
checkIfWordContainsLetters("google", ["a","o","o","g","g","l","e","x"]); // returns true
checkIfWordContainsLetters("google", ["a","o","g","g","l","e","x"]); // returns false