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Move string capital letters to front while maintaining the order (JavaScript)

This is my first post so I hope im doing this correctly.
I am taking a coding class and we were asked to make a piece of code that will ask for the input of a phrase, and will return in the console that phrase with the capital letters moved to the front, but still in the same order. Then print to the console this reordered phrase. (We aren't allowed to use arrays)
For example:
Inputting "HeLLoTherE" would return "HLLTEeoher"
However the problem is im having issues understanding how to write this code. How can I make the code select these capital letters and move them to the front? using .toUpperCase()? How can i make that select the letter and move it in front of the rest?
If someone could show me an example of how this is done and explain it a little i would greatly appreciate it :)

You might just start with a the most straight forward algorithm to get something working.
let value = "HeLLoTherE";
let result = "";
for (let char of value) {
if (char >= "A" && char <= "Z") {
result += char;
}
}
for (let char of value) {
if (char >= "a" && char <= "z") {
result += char;
}
}
console.log(result);
You could then consolidate the 2 loops by combining the conditions.
let value = "HeLLoTherE";
let upper = "";
let lower = "";
for (let char of value) {
if (char >= "A" && char <= "Z") {
upper += char;
} else if (char >= "a" && char <= "z") {
lower += char;
}
}
console.log(upper + lower);
Another way of solving this would be to use regex.
var value = "HeLLoTherE";
var upper = value.replace(/[^A-Z]*/g, "");
var lower = value.replace(/[^a-z]*/g, "");
console.log(upper + lower);

Well, you are not able to use arrays, which makes it a little bit difficult, however you can still do sommething.
Although I'm using a for loop, I'm not actually using arrays. Since strings allows the [] operator, you can use an index to select each character of the string and check if it's lowercase or uppercase.
In addition, you said you need to mantain the order of uppercase letters, so you couldn't just do newStr = upper + newStr, because it would revert the original order. So, I used the string.prototype.substring() to insert the uppercase character where it should be.
const str = "HeLLoTherE";
const moveUpperToFront = (target) => {
// Strings are immutable in js, so you cannot move one character
// to the front without using a new string.
let newStr = "";
// Number of uppercase letters that appeared.
// It's necessary because you need to mantain the original order
let upperNumber = 0;
// Iterate each character from beginning
for (let i = 0; i < str.length; ++i) {
// Is there an uppercase letter?
if (str[i].charCodeAt() >= 65 && str[i].charCodeAt() <= 90) {
newStr =
newStr.substring(0, upperNumber) +
str[i] +
newStr.substring(upperNumber, newStr.length);
++upperNumber;
}
// No uppercase letter?
else
newStr += str[i];
}
return newStr;
};
console.log(moveUpperToFront(str));

Following a solution which uses a for...of loop to iterate the input. It splits the input into capital and lowercase literals and then merges back together:
const exampleLiteral = 'HeLLoTherE';
const isUppercase = (literal) => literal === literal.toUpperCase() && literal !== literal.toLowerCase();
const prefixCapitalLetters = (literal) => {
let capitalLetters = '';
let lowerLetters = '';
for (let letter of literal) {
if(isUppercase(letter)) {
capitalLetters = capitalLetters.concat(letter);
continue;
}
lowerLetters = lowerLetters.concat(letter);
};
return capitalLetters+lowerLetters;
}
console.log(prefixCapitalLetters(exampleLiteral));

This is really not a very hard problem:
function rearrange(str) {
let result = "";
for (let c of str)
if (c >= 'A' && c <= 'Z')
result += c;
for (let c of str)
if (c < 'A' || c > 'Z')
result += c;
return result;
}
console.log(rearrange("Hello World, It Is A Beautiful Morning!"));
Find the upper-case characters, and add them to a result string. Then go back and find the other characters, and add them at the end. By looping through without any sorting, just simple iteration from start to finish, the order is preserved (other than the upper-case stuff).
The truly hard part of this would be coming up with a way to detect "upper-case" letters across all of Unicode. Some languages (well, orthographies) don't have the concept at all. JavaScript has ways that are more and less convenient to deal with that, but I suspect for the classroom material the OP has available so far, given the nature of the original question, such regex trickery would probably be inappropriate for an answer.

This answer tries to achieve the desired objective without using "arrays". It does use back-ticks, but that can be replaced with a simple string-concatenation if required.
Code Snippet
// move upper-case letters while
// keeping relative order same
const capsWithOrder = str => {
// initialize result variables
let capsOnly = "", restAll = "";
// iterate over the given string input
for (let i = 0; i < str.length; i++) {
// if character at index "i" is upper-case
// then, concatenate character to "capsOnly"
// else, concatenate to "restAll"
if (str[i] === str[i].toUpperCase()) capsOnly += str[i];
else restAll += str[i];
};
// after iterating over all characters in string-input
// return capsOnly concatenated with restAll
return `${capsOnly}${restAll}`;
};
console.log(capsWithOrder("HeLLoTherE"));
Explanation
Inline comments added in the snippet above.

Something like this
const string1 = 'HeLLoTherE'
const transform = string => {
const lower = string.split('').filter(c => c.charCodeAt() > 'a'.charCodeAt())
const upper = string.split('').filter(c => c.charCodeAt() < 'Z'.charCodeAt())
return [...upper, ...lower].join('')
}
console.log(transform(string1))

I think that must be work.
const sort = [
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split(''),
'abcdefghijklmnopqrstuvwxyz'.split('')
]
function listByValue(string) {
string = [...string];
let ret = [];
for (let i in sort)
ret = [...ret,...string.filter(e=>sort[i].includes(e))];
return ret.join('')
}

How make a more concise code using the For statement?

I'm new in StackOverflow and JavaScript, I'm trying to get the first letter that repeats from a string considering both uppercase and lowercase letters and counting and obtaining results using the for statement. The problem is that the form I used is too long Analyzing the situation reaches such a point that maybe you can only use a "For" statement for this exercise, which I get to iterate, but not with a cleaner and reduced code has me completely blocked, this is the reason why I request help to understand and continue with the understanding and use of this sentence. In this case, the result was tested in a JavaScript script inside a function and 3 "For" sentences obtaining quite positive results, but I can not create it in 1 only For (Sorry for my bad english google translate)
I making in HTML with JavasScript
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
var contendor = [];
var calc = [];
var mycalc = 0;
letter = letter.toUpperCase()
console.log(letter)
function repeats(){
for (var i = 0; i < letter.length; i++) {
if (contendor.includes(letter[i])) {
}else{
contendor.push(letter[i])
calc.push(0)
}
}
for (var p = 0; p < letter.length; p++) {
for (var l = 0; l < contendor.length; l++) {
if (letter[p] == contendor[l]) {
calc [l]= calc [l]+1
}
}
}
for (var f = 0; f < calc.length; f++) {
if ( calc[f] > calc[mycalc]) {
mycalc = f
}
}
}
repeats()
console.log("The most repeated letter its: " + contendor[mycalc]);
I Expected: A result with concise code

It would probably be a lot more concise to use a regular expression: match a character, then lookahead for more characters until you can match that first character again:
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
const firstRepeatedRegex = /(.)(?=.*\1)/;
console.log(letter.match(firstRepeatedRegex)[1]);
Of course, if you aren't sure whether a given string contains a repeated character, check that the match isn't null before trying to extract the character:
const input = 'abcde';
const firstRepeatedRegex = /(.)(?=.*\1)/;
const match = input.match(firstRepeatedRegex);
if (match) {
console.log(match[0]);
} else {
console.log('No repeated characters');
}
You could also turn the input into an array and use .find to find the first character whose lastIndexOf is not the same as the index of the character being iterated over:
const getFirstRepeatedCharacter = (str) => {
const chars = [...str];
const char = chars.find((char, i) => chars.lastIndexOf(char) !== i);
return char || 'No repeated characters';
};
console.log(getFirstRepeatedCharacter('abcde'));
console.log(getFirstRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));
If what you're actually looking for is the character that occurs most often, case-insensitive, use reduce to transform the string into an object indexed by character, whose values are the number of occurrences of that character, then identify the largest value:
const getMostRepeatedCharacter = (str) => {
const charsByCount = [...str.toUpperCase()].reduce((a, char) => {
a[char] = (a[char] || 0) + 1;
return a;
}, {});
const mostRepeatedEntry = Object.entries(charsByCount).reduce((a, b) => a[1] >= b[1] ? a : b);
return mostRepeatedEntry[0];
};
console.log(getMostRepeatedCharacter('abcde'));
console.log(getMostRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));

If the first repeated character is what you want, you can push it into an array and check if the character already exists
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
This will return the first repeating character if it exists, or will return -1.
Working
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
console.log(getFirstRepeating("SYAHSVCXCyXSssssssyBxAVMZsXhZV"))

Have you worked with JavaScript objects yet?
You should look into it.
When you loop through your string
let characters = "hemdhdksksbbd";
let charCount = {};
let max = { count: 0, ch: ""}; // will contain max
// rep letter
//Turn string into an array of letters and for
// each letter create a key in the charcount
// object , set it to 1 (meaning that's the first of
// that letter you've found) and any other time
// you see the letter, increment by 1.
characters.split("").forEach(function(character)
{
if(!charCount[character])
charCount[character] = 1;
else
charCount[character]++;
}
//charCount should now contain letters and
// their counts.
//Get the letters from charCount and find the
// max count
Object.keys(charCount). forEach (function(ch){
if(max.count < charCount[ch])
max = { count: charCount[ch], ch: ch};
}
console.log("most reps is: " , max.ch);

This is a pretty terrible solution. It takes 2 loops (reduce) and doesn't handle ties, but it's short and complicated.
Basically keep turning the results into arrays and use array methods split and reduce to find the answer. The first reduce is wrapped in Object.entries() to turn the object back into an array.
let letter = Object.entries(
"SYAHSVCXCyXSssssssyBxAVMZsXhZV".
toUpperCase().
split('').
reduce((p, c) => {
p[c] = isNaN(++p[c]) ? 1 : p[c];
return p;
}, {})
).
reduce((p, c) => p = c[1] > p[1] ? c : p);
console.log(`The most repeated letter is ${letter[0]}, ${letter[1]} times.`);

Find smallest substring containing a given set of letters in a larger string

Say you have the following string:
FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNT
LDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFY
FFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQ
XBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR
AMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR
I'm trying to find the smallest substring containing the letters ABCDA.
I tried a regex approach.
console.log(str.match(/[A].*?[B].*?[C].*?[D].*?[A]/gm).sort((a, b) => a.length - b.length)[0]);
This works, but it only find strings where ABCDA appear (in that order). Meaning it won't find substring where the letters appear in a order like this: BCDAA
I'm trying to change my regex to account for this. How would I do that without using | and type out all the different cases?

You can't.
Let's consider a special case: Assume the letters you are looking for are A, A, and B. At some point in your regexp there will certainly be a B. However, the parts to the left and to the right of the B are independent of each other, so you cannot refer from one to the other. How many As are matched in the subexpression to the right of the B depends on the number of As being already matched in the left part. This is not possible with regular expressions, so you will have to unfold all the different orders, which can be many!
Another popular example that illustrates the problem is to match opening brackets with closing brackets. It's not possible to write a regular expression asserting that in a given string a sequence of opening brackets is followed by a sequence of closing brackets of the same length. The reason for this is that to count the brackets you would need a stack machine in contrast to a finite state machine but regular expressions are limited to patterns that can be matched using FSMs.

This algorithm doesn't use a regex, but found both solutions as well.
var haystack = 'FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGRAMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR';
var needle = 'ABCDA'; // the order of letters doesn't matter
var letters = {};
needle.split('').forEach(function(ch) {
letters[ch] = letters[ch] || 0;
letters[ch]++;
});
var shortestSubstringLength = haystack.length;
var shortestSubstrings = []; // storage for found substrings
var startingPos = 0;
var length;
var currentPos;
var notFound;
var letterKeys = Object.keys(letters); // unique leters
do {
lettersLeft = JSON.parse(JSON.stringify(letters)); // copy letters count object
notFound = false;
posStart = haystack.length;
posEnd = 0;
letterKeys.forEach(function(ch) {
currentPos = startingPos;
while (!notFound && lettersLeft[ch] > 0) {
currentPos = haystack.indexOf(ch, currentPos);
if (currentPos >= 0) {
lettersLeft[ch]--;
posStart = Math.min(currentPos, posStart);
posEnd = Math.max(currentPos, posEnd);
currentPos++;
} else {
notFound = true;
}
}
});
if (!notFound) {
length = posEnd - posStart + 1;
startingPos = posStart + 1; // starting position for next iteration
}
if (!notFound && length === shortestSubstringLength) {
shortestSubstrings.push(haystack.substr(posStart, length));
}
if (!notFound && length < shortestSubstringLength) {
shortestSubstrings = [haystack.substr(posStart, length)];
shortestSubstringLength = length;
}
} while (!notFound);
console.log(shortestSubstrings);

Maybe not as clear as using regex could be (well, for me regex are never really clear :D ) you can use brute force (not so brute)
Create an index of "valid" points of your string (those with the letters you want) and iterate with a double loop over it getting substrings containing at least 5 of those points, checking that they are valid solutions. Maybe not the most efficient way, but easy to implement, to understand, and probably to optimize.
var haystack="UGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR";
var needle="ABCD";
var size=haystack.length;
var candidate_substring="";
var minimal_length=size;
var solutions=new Array();
var points=Array();
for(var i=0;i<size;i++){
if(needle.indexOf(haystack[i])>-1) points.push(i);
}
var limit_i= points.length-4;
var limit_k= points.length;
for (var i=0;i<limit_i;i++){
for(var k=i;k<limit_k;k++){
if(points[k]-points[i]+1<=minimal_length){
candidate_substring=haystack.substr(points[i],points[k]-points[i]+1);
if(is_valid(candidate_substring)){
solutions.push(candidate_substring);
if(candidate_substring.length < minimal_length) minimal_length=candidate_substring.length;
}
}
}
}
document.write('<p>Solution length:'+minimal_length+'<p>');
for(var i=0;i<solutions.length;i++){
if(solutions[i].length<=minimal_length) document.write('<p>Solution:'+solutions[i]+'<p>');
}
function is_valid(candidate_substring){
//verify we've got all characters
for(var j=0;j<candidate_substring.length;j++){
if(candidate_substring.indexOf(needle.charAt(j))<0) return false;
}
//...and verify we have two "A"
if(candidate_substring.indexOf("A")==candidate_substring.lastIndexOf("A")) return false;
return true;
}

Just had this problem in an interview as a coding assignment and came up with another solution, (it's not as optimal as the one above but maybe it's easier to understand).
function MinWindowSubstring(strArr) {
const N = strArr[0];
const K = strArr[1];
const letters = {};
K.split('').forEach( (character) => {
letters[character] = letters[character] ? letters[character] + 1 : 1;
});
let possibleSequencesList = [];
const letterKeys = Object.keys(letters);
for(let i=0; i< N.length; i++) {
const char = N[i];
if (new String(letterKeys).indexOf(char) !== -1) {
// found a character in the string
// update all previus sequences
possibleSequencesList.forEach((seq) => {
if(!seq.sequenceComplete) {
seq[char] = seq[char]-1;
seq.lastIndex = i;
// check if sequence is complete
var sequenceComplete = true;
letterKeys.forEach( (letter) => {
if(seq[letter] > 0) {
sequenceComplete = false;
}
});
seq.sequenceComplete = sequenceComplete
}
})
// create a new sequence starting from it
const newSeq = {
startPoint: i,
lastIndex: i,
sequenceComplete: false,
...letters
}
newSeq[char] = newSeq[char]-1;
possibleSequencesList.push(newSeq);
}
}
// cleanup sequences
let sequencesList = possibleSequencesList.filter(sequence => sequence.sequenceComplete);
let output = [];
let minLength = N.length;
// find the smalles one
sequencesList.forEach( seq => {
if( (seq.lastIndex - seq.startPoint) < minLength) {
minLength = seq.lastIndex - seq.startPoint;
output = N.substring(seq.startPoint, seq.lastIndex + 1);
}
})
return output;
}

Getting out an expression

Let say I have these two examples
(A = 1) and ( B = 2)
(A = 1)(B = 2 ()).
I need a way to get the following array:
[(],[A][=][1],[)],[and],[(],[B],[=],[2],[)]
[(],[A][=][1],[)],[(],[B],[=],[2],[(],,[)][)]
What I tried to do is the following
Find the delimiters using the following function (in this case the delimiters are the space "" and any brackets ( or ) )
function findExpressionDelimeter (textAreaValue){
var delimiterPositions = [];
var bracesDepth = 0;
var squareBracketsDepth = 0;
var bracketsDepth = 0;
for (var i = 0; i < textAreaValue.length; i++) {
switch (textAreaValue[i]) {
case '(':
bracketsDepth++;
delimiterPositions.push(i);
break;
case ')':
bracketsDepth--;
delimiterPositions.push(i);
break;
case '[':
squareBracketsDepth++;
break;
case ']':
squareBracketsDepth--;
break;
default:
if (squareBracketsDepth == 0 && textAreaValue[i] == ' ') {
delimiterPositions.push(i);
}
}
}
return delimiterPositions;
}
Then I tried to loop trough the values returned and extract the values using substring. The issue is that when I have a ( or ) I need to get the next substring as well as the bracket. This is where I am stuck.
function getTextByDelimeter(delimiterPositions, value) {
var output = [];
var index = 0;
var length = 0;
var string = "";
for (var j = 0; j < delimiterPositions.length; j++) {
if (j == 0) {
index = 0;
} else {
index = delimiterPositions[j - 1] + 1;
}
length = delimiterPositions[j];
string = value.substring(index, length);
output.push(string);
}
string = value.substring(length, value.length);
output.push(string);
return output;
}
Any help would be appreciated.

You could just match the tokens you are interested in:
var str = "(A = 1) and ( B = 2)";
var arr = str.match(/[()]|[^()\s]+/g);
Result:
["(", "A", "=", "1", ")", "and", "(", "B", "=", "2", ")"]
The regex with some comments:
[()] # match a single character token
| # or
[^()\s]+ # match everything else except spaces
If you would like to add more single character tokens, like for example a =, just add it to both character classes. Ie: [()=]|[^()=\s]+

What you want to do is a lexical analyser.
Regular expressions won't allow you to parse a language (a mathematical expression is one). The tree decomposition of the formula cannot be done with it.
However, regex can allow you to discriminate tokens. This is usually done by reading the stream of character. Once you've detect a lexeme, you generate the token.
If you want to check the validity of the formula, or compute the value: you need a parser (semantic analyser). This can't be done using regex.

The similar question with the answer is here.
You can split your string(string.split('')) And then delete whitespaces from array or just check if array[i] != ' ' before your switch block.

pass $` value to associated parameter function of replace

I have an expression say
log(1,3)+4,5+max(7,8,9)
where comma is being used two ways.
1- In "log(1,3)+4,5" comma is being used in place of dot(.) or decimal sign.i.e. "log(1,3)+4,5" is equivalent to "log(1.3)+4.5".
2- In max(7,8,9) it is being used as number separator. i.e. this outcome of this is 9 ; the maximum number.
My problem is to substitute comma; which is being used as decimal separator; with decimal but this should not affect max(7,8,9). i.e. I need to convert above expression to
log(1.3)+4.5+max(7,8,9)
What I tried-
function substitute(expr) {
expr.replace(/,/g, function ($`) {
/*some processing here to decide whether comma to be substituted with dot or not.On that basis I will return either dot or comma.*/
}
But how can I pass $` value to associated function
or
Is it possible to do this in javascript.
expr.replace(/,/g,function ($`) {
if yes then how?

Your language is ambiguous.
max(8,1,8,2)
Does this return 8, 8,1 or 8,2?
Your language also doesn't look regular, so you can't parse it with a regular expression, you need the context. If something like this is allowed:
max(1,max(2,3)) // 3?
Assuming you can get rid of the ambiguity, you could write a parser to do the context detection.

This could be a solution :
function myFilter(string) {
// save all functions and signs
var functions = [];
var regExp = /[+,-]max\(([^\)]+)\)/;
matches = true;
while (matches !== null) {
var matches = regExp.exec(string);
if (matches !== null) {
functions.push(matches[0]);
string = string.replace(matches[0], '');
}
}
// replace all remaining commas with dots
string = string.replace(/,/g , ".");
for (i in functions) {
string += functions[i];
}
return string;
}
var s = '1,3+4,5+max(7,8,9)-max(2,3,5)';
var filteredString = myFilter(s);
jsFiddle Demo
This currently works with multiple max functions but only + and - signs. It could be improved with *, / and more... You will have to find the good regex.

Try the below using Javascript. Hope this helps you in logic.
DEMO HERE
var value = "log(1,3)-4,5+max(7,8,9)";
var val = '';
var splitValue, appendSym;
if (value.indexOf("+") != -1)
{
splitValue = value.split("+");
appendSym = "+";
}
else if(value.indexOf("-") != -1)
{
splitValue = value.split("-");
appendSym = "-";
}
else if(value.indexOf("*") != -1)
{
splitValue = value.split("*");
appendSym = "*";
}
else
{
splitValue = value.split("/");
appendSym = "/";
}
var length = splitValue.length;
for (var i = 0; i < length; i++) {
if (val) val += appendSym;
var strrep = splitValue[i].replace(/,/g,".");
if (splitValue[i].indexOf("max") != -1 || splitValue[i].indexOf("min") != -1)
{
val+=splitValue[i];
}
else
{
val+=strrep;
}
}
alert(val);
The output for the above code is log(1.3)-4.5+max(7,8,9)