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Return all possible combinations of array with optional strings

Lets say I have an array keys = ["the?", "orange", "van", "s?"], with '?' at the end of strings to represent that it is optional.
I want a function in javascript generateCombinations(keys) that returns the possible combinations such as :
[["orange","van"],["the","orange","van"],["orange","van","s"],["the","orange","van","s"]]
One possible way of removing '?' is to simply do a replace("?',"").
I have a feeling it might require a recursive function, which I am not yet quite strong in. Help is appreciated!
So far I've tried this:
function isOptionalKey(key) {
return key.endsWith('?');
}
function hasOptionalKey(keys) {
return keys.some(isOptionalKey);
}
function stripOptionalSyntax(key) {
return key.endsWith('?') ? key.slice(0, -1) : key;
}
function generateCombinations(keys) {
if (keys.length === 1) {
return keys;
}
const combinations = [];
const startKey = keys[0];
const restKeys = keys.slice(1);
if (hasOptionalKey(restKeys)) {
const restCombinations = isOptionalKey(startKey)
? generateCombinations(restKeys)
: restKeys;
if (isOptionalKey(startKey)) {
combinations.push(restCombinations);
}
combinations.push(
restCombinations.map((c) => [stripOptionalSyntax(startKey), ...c])
);
} else {
if (isOptionalKey(startKey)) {
combinations.push(restKeys);
}
combinations.push([stripOptionalSyntax(startKey), ...restKeys]);
}
return combinations;
}

You could take a recursive approach by using only the first item of the array and stop if the array is empty.
const
getCombinations = array => {
if (!array.length) return [[]];
const
sub = getCombinations(array.slice(1)),
optional = array[0].endsWith('?'),
raw = optional ? array[0].slice(0, -1) : array[0],
temp = sub.map(a => [raw, ...a]);
return optional
? [...temp, ...sub]
: temp;
};
keys = ["the?", "orange", "van", "s?"],
result = getCombinations(keys);
console.log(result.map(a => a.join(' ')));

how can i count chars from word which in the array?

i have an array ["academy"] and i need count chars from the string in the array.
output:
a:2
c:1
d:1
e:1
m:1
y:1
like this
i tried two for loops
function sumChar(arr){
let alph="abcdefghijklmnopqrstuvxyz";
let count=0;
for (const iterator of arr) {
for(let i=0; i<alph.length; i++){
if(iterator.charAt(i)==alph[i]){
count++;
console.log(`${iterator[i]} : ${count}`);
count=0;
}
}
}
}
console.log(sumChar(["abdulloh"]));
it works wrong
Output:
a : 1
b : 1
h : 1
undefined

Here's a concise method. [...new Set(word.split(''))] creates an array of letters omitting any duplicates. .map takes each letter from that array and runs it through the length checker. ({ [m]: word.split(m).length - 1 }) sets the letter as the object key and the word.split(m).length - 1is a quick way to determine how many times that letter shows up.
const countLetters = word => (
[...new Set(word.split(''))].map(m => ({
[m]: word.split(m).length - 1
})))
console.log(countLetters("academy"))

You can check the occurrences using regex also. in this i made a method which checks for the character in the string. Hope it helps.
word: string = 'abcdefghijklkmnopqrstuvwxyzgg';
charsArrayWithCount = {};
CheckWordCount(): void {
for(var i = 0;i < this.word.length; i++){
if(this.charsArrayWithCount[this.word[i]] === undefined){
this.charsArrayWithCount[this.word[i]] = this.charCount(this.word, this.word[i]);
}
}
console.log(this.charsArrayWithCount);
}
charCount(string, char) {
let expression = new RegExp(char, "g");
return string.match(expression).length;
}

You can simply achieve this requirement with the help of Array.reduce() method.
Live Demo :
const arr = ["academy"];
const res = arr.map(word => {
return word.split('').reduce((obj, cur) => {
obj[cur] = obj[cur] ? obj[cur] + 1 : 1
return obj;
}, {});
});
console.log(res);

I think this is the simplest:
const input = 'academy';
const res = {};
input.split('').forEach(a => res[a] = (res[a] ?? 0) + 1);
console.log(res);

I have a array of string have to find all the common character present from all strings

I have a array of string.
let arr=["robin","rohit","roy"];
Need to find all the common character present in all the strings in array.
Output Eg: r,o
I have tried to create a function for above case with multiple loops but i want to know what should be the efficient way to achive it.

Here's a functional solution which will work with an array of any iterable value (not just strings), and uses object identity comparison for value equality:
function findCommon (iterA, iterB) {
const common = new Set();
const uniqueB = new Set(iterB);
for (const value of iterA) if (uniqueB.has(value)) common.add(value);
return common;
}
function findAllCommon (arrayOfIter) {
if (arrayOfIter.length === 0) return [];
let common = new Set(arrayOfIter[0]);
for (let i = 1; i < arrayOfIter.length; i += 1) {
common = findCommon(common, arrayOfIter[i]);
}
return [...common];
}
const arr = ['robin', 'rohit', 'roy'];
const result = findAllCommon(arr);
console.log(result);

const arr = ["roooooobin","rohit","roy"];
const commonChars = (arr) => {
const charsCount = arr.reduce((sum, word) => {
const wordChars = word.split('').reduce((ws, c) => {
ws[c] = 1;
return ws;
}, {});
Object.keys(wordChars).forEach((c) => {
sum[c] = (sum[c] || 0) + 1;
});
return sum;
}, {});
return Object.keys(charsCount).filter(key => charsCount[key] === arr.length);
}
console.log(commonChars(arr));

Okay, the idea is to count the amount of times each letter occurs but only counting 1 letter per string
let arr=["robin","rohit","roy"];
function commonLetter(array){
var count={} //object used for counting letters total
for(let i=0;i<array.length;i++){
//looping through the array
const cache={} //same letters only counted once here
for(let j=0;j<array[i].length;j++){
//looping through the string
let letter=array[i][j]
if(cache[letter]!==true){
//if letter not yet counted in this string
cache[letter]=true //well now it is counted in this string
count[letter]=(count[letter]||0)+1
//I don't say count[letter]++ because count[letter] may not be defined yet, hence (count[letter]||0)
}
}
}
return Object.keys(count)
.filter(letter=>count[letter]===array.length)
.join(',')
}
//usage
console.log(commonLetter(arr))

No matter which way you choose, you will still need to count all characters, you cannot get around O(n*2) as far as I know.
arr=["robin","rohit","roy"];
let commonChars = sumCommonCharacters(arr);
function sumCommonCharacters(arr) {
data = {};
for(let i = 0; i < arr.length; i++) {
for(let char in arr[i]) {
let key = arr[i][char];
data[key] = (data[key] != null) ? data[key]+1 : 1;
}
}
return data;
}
console.log(commonChars);

Here is a 1 liner if anyone interested
new Set(arr.map(d => [...d]).flat(Infinity).reduce((ac,d) => {(new RegExp(`(?:.*${d}.*){${arr.length}}`)).test(arr) && ac.push(d); return ac},[])) //{r,o}

You can use an object to check for the occurrences of each character. loop on the words in the array, then loop on the chars of each word.
let arr = ["robin","rohit","roy"];
const restWords = arr.slice(1);
const result = arr[0].split('').filter(char =>
restWords.every(word => word.includes(char)))
const uniqueChars = Array.from(new Set(result));
console.log(uniqueChars);

How could I find intersections in a string array with a variable number of elements?

I have a function that receives an array composed of numerical, comma separated strings as input, then finds the intersectional numbers in those strings and returns a string of similarly comma separated numbers, with no spaces, containing those intersections. If there are no intersections between the two elements, the function will return false.
What I want is to optimize the function so that it can work with a string array that may have more than just two elements. Is that possible? If so, could I have some sort of guideline of where to start looking for answers?
Currently, this is what I have.
function LocateIntersection(strArr) {
let arrHalf1 = strArr[0].split(", ");
let arrHalf2 = strArr[1].split(", ");
let interArr = arrHalf1.filter(value => arrHalf2.includes(value));
let result = interArr.join();
if (result) {
return result;
} else {
return false;
}
}

My answer is a little flawed, but it should meet your requirements
function LocateIntersection(strArr) {
const AllArrHalf = strArr.map((value) => value.split(', ')).sort((a, b) => b.length - a.length);
const lastArrHalf = AllArrHalf[AllArrHalf.length - 1];
let interArr = [];
AllArrHalf.forEach((value, index) => {
if (index !== AllArrHalf.length - 1) {
interArr.push(lastArrHalf.filter(value1 => value.includes(value1)))
}
})
if (interArr.length > 1) {
let result = interArr.map(value => value.join(', '));
LocateIntersection(result);
} else if (interArr.length === 1) {
result = interArr.join();
console.log(result);
}
}
LocateIntersection(['a, b, c', 'a, b', 'a, b'])

You can try this.
const intersection = (arr1, arr2) => {
return arr2.filter(element => arr1.includes(element));
}
const getIntersection = (stringArray, prevResult) => {
const array1 = prevResult || stringArray[0].split(', ');
const array2 = stringArray.shift().split(', ');
const result = intersection(array1, array2);
console.log(`result : `, result)
if(result.length > 0 && stringArray.length > 0) {
return getIntersection(stringArray, result);
}
return result;
}
const input = ['1, 2','1, 3, 3, 3','123, 222','1, 1, 1','1','3, 2, 3, 1'];
const result = getIntersection(input);
console.log('final Result:',result);

Convert string into key-value pairs in an array

I have a string:
var rrule = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
I want to convert this string to key-> value pairs in an array.
[
dtstart: 20190514T111500Z,
freq: daily,
interval: 1
]
I know I can take the string and split it based on the semicolon:
var array = rrule.split(";");
... but this leaves me with an array like this:
[
"DTSTART=20190514T111500Z",
"FREQ=DAILY",
"INTERVAL=1"
]
I guess I need another step to map out the keys/values, but I get lost at this point.
Ideally, for the string I want to be able to easily access what dtstarts equals, what interval equals, what other variables equal and so on.

let str = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
let obj = {};
for (let entry of str.split(";")) {
let pair = entry.split("=");
obj[pair[0]] = pair[1];
}
console.log(obj);

You already know how to split on the ; to get an array, from there you can just aggregate (using reduce) to get an object:
var rrule = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
var result = rrule.split(";").reduce( (obj,item) => {
let [key,value] = item.split("=");
obj[key] = value;
return obj;
},{});
console.log(result["DTSTART"])
console.log(result["FREQ"])
console.log(result["INTERVAL"])

You were correct to start with split first, this would then return you an array of strings.
To easily convert them, just use map, to return the split the single strings once more, and then return an object based on the property name you would like to give it and it's value
function createKeyValuePairFromString( str ) {
return str.split(';').map( item => {
const splitted = item.split('=');
return { [splitted[0]]: splitted[1] };
});
}
console.log( createKeyValuePairFromString("DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1") );

Use array created and split it again with =
function convertToObject(cookieString) {
const cookieObj = {};
if (!cookieString && typeof cookieString !== 'string') return cookieObj;
const arr = cookieString.split(';');
arr.forEach(record => {
if (record.includes('=')) {
const [key, value] = record.split('=');
cookieObj[key.trim()] = value;
}
});
return cookieObj;
}

You can use it like the code below:
var rrule = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
let finalObj = {};
rrule.split(';').forEach(i => finalObj[i.split('=')[0]] = i.split('=')[1]);
console.log('finalObj',finalObj);
Here I'm first splitting with ';' so consider the first item to be DTSTART=20190514T111500Z Then on splitting with = I get finalObject['DTSTART'] = 20190514T111500Z

Using forEach()
let str = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
let obj = {};
let strArr = str.split(';')
strArr.forEach((str) => {
let [key, value] = str.split('=')
obj[key] = value;
});
console.log(obj);

Here's a fairly simple version, returning an object, not an array:
const toObj = str => str
.split (';')
.map ( s => s .split ('=') )
.reduce ( (a, [k, v]) => ({...a, [k]: v}), {} )
let str = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
console.log (
toObj(str)
)
One of the reasons I like the library is that we can write this sort of logic more simply. In Ramda (disclaimer: I'm one of the authors), it might look like this:
const toObj = pipe ( split (';'), map (split ('=') ), fromPairs)
let str = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
console.log (
toObj(str)
)
<script src="https://bundle.run/ramda#0.26.1"></script><script>
const {pipe, split, map, fromPairs} = ramda; </script>

var str = "DTSTART=20190514T111500Z;FREQ=DAILY;INTERVAL=1";
// string splitting rule
const rule = (string, delimiter) => string.split(delimiter);
const result = rule(str, ';').reduce((acc, s) => {
const [key, value] = rule(s, '=');
acc[key] = value;
return acc;
}, {});
console.log(result);