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How to find the longest sequence of "broken" characters using Regex in JS?

I'm trying to find the longest sequence of "broken"(different from letter, digit, space) characters in this sentence:
'Tempera####&&#^ na ##$ata 23 grad#%&.'
I really want to do it using Regex, but I'm not that familiar with the usage of Regex in JS. This is the description of the problem: https://pastebin.com/U6Uukc4b
I watched a lot of tutorials, also read bunch of articles and this is what I came up with:
let input = [
'Tempera####&&#^ na ##$ata 23 grad#%&.'
];
let print = this.print || console.log;
let gets = this.gets || ((arr, index) => () => arr[index++])(input, 0);
let message = gets();
//different from letter, digit, space
let counter = 1;
let allWords = /[a-z1-9\s]/gi;
for (let i = 0; i < message.length; i++) {
let current = message[i];
// allWords.lastIndex = 0;
let isExisting = allWords.test(current);
if (current === '.') {
break;
}
if (isExisting) {
counter = 1;
} else {
counter++;
}
}
print(counter)
Here the answer should be 8 , because we have 8 consecutive symbols inside the word "Tempera####&&#^" , which are not letter, digit or space.
Unfortunately this doesn't work. I'm still trying to understand where is my mistake, so I'll be very thankful if someone can "show me the way".

Use a regular expression that matches the opposite ([^.....]) and as many times as possible (+). Then check the size of all matches and pick the longest length. As there is a special role for the point (which apparently is always at the end), consider it as not broken.
let allWords = /[^a-z1-9\s.]+/gi;
let input = [
'Tempera####&&#^ na ##$ata 23 grad#%&.'
];
for (let message of input) {
let results = message.match(allWords) ?? [];
let counter = Math.max(0, ...results.map(({length}) => length));
console.log("longest", counter);
}

const input = 'Tempera####&&#^ na ##$ata 23 grad#%&.';
function findLongestSubstring(input) {
let longest = '';
let current = '';
for (let i = 0; i < input.length; i++) {
if (input[i].match(/[^a-zA-Z\d\s]/)) {
current += input[i];
} else {
if (current.length > longest.length) {
longest = current;
}
current = '';
}
}
return longest.length;
}
console.log(findLongestSubstring(input));

Why won't this iterate?

I am trying to make this Pig Latin function (I just started coding 3 weeks ago, so go easy on me), and I can't figure out why I can't get the array made from .split(' ') and then iterated through to join back again. In the output I only get the first word. The code is below:
function pigLatin(str) {
let str1 = str.split(' ')
for (let i = 0; i < str1.length; i++) {
if (str1[i].length <= 1) {
return str1[i];
}
else {
let first = str1[i].substring(0,1);
let word = str1[i].substring(1);
str = word + first + 'ay';
return str
}
}
}
console.log(pigLatin("This is a test"));
Keep in mind that I was considering adding regex and more else if statements, but I can't even get this to work yet. Any help is greatly appreciated.

You're returning too early. You should be adding each word to an array, and at the end of your loop you should concatenate the words in the array to form a new string which you should return. See my comments for how I altered your code:
function pigLatin(str) {
let r = [] // The array to build
let str1 = str.split(' ')
for (let i = 0; i < str1.length; i++) {
if (str1[i].length <= 1) {
r.push( str1[i] ); // Add to end of array
}
else {
let first = str1[i].substring(0,1);
let word = str1[i].substring(1);
str = word + first + 'ay';
r.push(str) // Add to end of array
}
}
return r.join(' ') // Join strings in array and return new string
}
console.log(pigLatin("This is a test"));

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.`);

Removing zeros after comma based on maximum consequent zeros

I have a page with a grid where user's numbers get saved. It has a following pattern - every number ends with 3 digits after comma. It doesn't look nice, when for example user's input is
123,450
123,670
123,890
It's much better to have just 2 numbers after comma, because last 0 is absolutely meaningless and redundant.
The way it still should have 3 digits is only if at least one element in an array doesn't end up with 0
For example:
123,455
123,450
123,560
In this case 1st element of the array has the last digit not equal to 0 and hence all the elements should have 3 digits. The same story with 2 or 1 zeros
Zeros are redundant:
123,30
123,40
123,50
Zeros are necessary:
123,35
123,40
123,50
The question is how can I implement it programatically? I've started like this:
var zeros2Remove = 0;
numInArray.forEach(function(item, index, numInArray)
{
var threeDigitsAfterComma = item.substring(item.indexOf(',') + 1);
for(var j = 2; j <= 0; j--)
{
if(threeDigitsAfterComma[j] == 0)
{
zeros2Remove =+ 1;
}
else //have no idea what to do..
}
})
Well in my implementation I don't know how to do it since I have to iterate through every element but break it if at least 1 number has a last digit equal to zero.. In order to do that I have to break outer loop, but don't know how and I'm absolutely sure that I don't have to...

I think the following code what you are looking for exactly , please manipulate numbers and see the changes :
var arr = ["111.3030", "2232.0022", "3.001000", "4","558.0200","55.00003000000"];
var map = arr.map(function(a) {
if (a % 1 === 0) {
var res = "1";
} else {
var lastNumman = a.toString().split('').pop();
if (lastNumman == 0) {
var m = parseFloat(a);
var res = (m + "").split(".")[1].length;
} else {
var m = a.split(".")[1].length;
var res = m;
}
}
return res;
})
var maxNum = map.reduce(function(a, b) {
return Math.max(a, b);
});
arr.forEach(function(el) {
console.log(Number.parseFloat(el).toFixed(maxNum));
});

According to MDN,
There is no way to stop or break a forEach() loop other than by throwing an exception. If you need such behavior, the forEach() method is the wrong tool. Use a plain loop or for...of instead.
If you convert your forEach loop to a for loop, you can break out of it with a label and break statement:
// unrelated example
let i;
let j;
outerLoop:
for (i = 2; i < 100; ++i) {
innerLoop:
for (j = 2; j < 100; ++j) {
// brute-force prime factorization
if (i * j === 2183) { break outerLoop; }
}
}
console.log(i, j);
I gave you an unrelated example because your problem doesn't need nested loops at all. You can find the number of trailing zeroes in a string with a regular expression:
function getTrailingZeroes (str) {
return str.match(/0{0,2}$/)[0].length;
}
str.match(/0{0,2}$/) finds between 0 and 2 zeroes at the end of str and returns them as a string in a one-element array. The length of that string is the number of characters you can remove from str. You can make one pass over your array of number-strings, breaking out when necessary, and use Array.map as a separate truncation loop:
function getShortenedNumbers (numInArray) {
let zeroesToRemove = Infinity;
for (const str of numInArray) {
let candidate = getTrailingZeroes(str);
zeroesToRemove = Math.min(zeroesToRemove, candidate);
if (zeroesToRemove === 0) break;
}
return numInArray.map(str => str.substring(0, str.length - zeroesToRemove);
}
All together:
function getTrailingZeroes (str) {
return str.match(/0{0,2}$/)[0].length;
}
function getShortenedNumbers (numInArray) {
let zeroesToRemove = Infinity;
for (const str of numInArray) {
let candidate = getTrailingZeroes(str);
zeroesToRemove = Math.min(zeroesToRemove, candidate);
if (zeroesToRemove === 0) break;
}
return numInArray.map(str => str.substring(0, str.length - zeroesToRemove));
}
console.log(getShortenedNumbers(['123,450', '123,670', '123,890']));
console.log(getShortenedNumbers(['123,455', '123,450', '123,560']));

This solution might seem a little cumbersome but it should work for all possible scenarios. It should be easy enough to make always return a minimal number of decimals places/leading zeros.
I hope it helps.
// Define any array
const firstArray = [
'123,4350',
'123,64470',
'123,8112390',
]
const oneOfOfYourArrays = [
'123,30',
'123,40',
'123,50',
]
// Converts 123,45 to 123.45
function stringNumberToFloat(stringNumber) {
return parseFloat(stringNumber.replace(',', '.'))
}
// For 123.45 you get 2
function getNumberOfDecimals(number) {
return number.split('.')[1].length;
}
// This is a hacky way how to remove traling zeros
function removeTralingZeros(stringNumber) {
return stringNumberToFloat(stringNumber).toString()
}
// Sorts numbers in array by number of their decimals
function byNumberOfValidDecimals(a, b) {
const decimalsA = getNumberOfDecimals(a)
const decimalsB = getNumberOfDecimals(b)
return decimalsB - decimalsA
}
// THIS IS THE FINAL SOLUTION
function normalizeDecimalPlaces(targetArray) {
const processedArray = targetArray
.map(removeTralingZeros) // We want to remove trailing zeros
.sort(byNumberOfValidDecimals) // Sort from highest to lowest by number of valid decimals
const maxNumberOfDecimals = processedArray[0].split('.')[1].length
return targetArray.map((stringNumber) => stringNumberToFloat(stringNumber).toFixed(maxNumberOfDecimals))
}
console.log('normalizedFirstArray', normalizeDecimalPlaces(firstArray))
console.log('normalizedOneOfOfYourArrays', normalizeDecimalPlaces(oneOfOfYourArrays))

Try this
function removeZeros(group) {
var maxLength = 0;
var newGroup = [];
for(var x in group) {
var str = group[x].toString().split('.')[1];
if(str.length > maxLength) maxLength = str.length;
}
for(var y in group) {
var str = group[y].toString();
var substr = str.split('.')[1];
if(substr.length < maxLength) {
for(var i = 0; i < (maxLength - substr.length); i++)
str += '0';
}
newGroup.push(str);
}
return newGroup;
}
Try it on jsfiddle: https://jsfiddle.net/32sdvzn1/1/
My script checks the length of every number decimal part, remember that JavaScript removes the last zeros in a decimal number, so 3.10 would be 3.1, so the length is less when there is a number with zeros in the end, in this case we just add a zero to the number.
Update
I've updated the script, the new version adds as much zeros as the different between the max decimal length and the decimal length of the analyzed number.
Example
We have: 3.11, 3.1423, 3.1
The max length would be: 4 (1423)
maxLenght (4) - length of .11 (2) = 2
We add 2 zeros to 3.11, that will become 3.1100

I think you can start out assuming you will remove two extra zeros, and loop through your array looking for digits in the last two places. With the commas, I'm assuming your numArray elements are strings, all starting with the same length.
var numArray = ['123,000', '456,100', '789,110'];
var removeTwo = true, removeOne = true;
for (var i = 0; i < numArray.length; i++) {
if (numArray[i][6] !== '0') { removeTwo = false; removeOne = false; }
if (numArray[i][5] !== '0') { removeTwo = false; }
}
// now loop to do the actual removal
for (var i = 0; i < numArray.length; i++) {
if (removeTwo) {
numArray[i] = numArray[i].substr(0, 5);
} else if (removeOne) {
numArray[i] = numArray[i].substr(0, 6);
}
}

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;
}