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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);

Find occurance of characters in array of object using js

I have array of object with few words I want to know occurance of each word and push into key value pair format
let words = ["aabbbc", "dddeeef", "gghhhii"]
Output
[{a:2, b:3,c:1}, {d:3,e:3,f:1}, {g:2,h:3:i:2}]

This is a classic map and reduce task where one maps the array of strings and for each string creates the character-specific counter-statistics while reducing the string's character-sequence (...split('').reduce( ... )) and by programmatically building an object which counts/totals each character's occurrence.
console.log(
["aabbbc", "dddeeef", "gghhhii"]
.map(string =>
string
.split('')
.reduce((result, char) => {
result[char] = (result[char] ?? 0) + 1;
return result;
}, {})
)
)
// [{a:2, b:3,c:1}, {d:3,e:3,f:1}, {g:2,h:3:i:2}]
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let words = ["aabbbc", "dddeeef", "gghhhii"]
const occurences = (w) => {
const obj = {};
for (const c of w) {
if (obj[c] === undefined) obj[c] = 0;
obj[c]++;
}
return obj;
}
const arr = words.map(w => occurences(w));
console.log(arr)

Trying to keep it easy and readable:
const words = ['aabbbc', 'dddeeef', 'gghhhii']
const output = []
for (const word of words) {
const result = {}
for (const letter of word) {
result[letter] = result[letter] || 0
result[letter]++
}
output.push(result)
}
console.log({ output })

let words = ["aabbbc", "ddeeef", "ghhhii"]
let newArr = []
words.forEach((e,index)=>{
e.split('').forEach(n=>{
if(!newArr[index]){
newArr[index]={}
}
if(!newArr[index][n]){
newArr[index][n]=0
}
(newArr[index][n]>=0) && ++newArr[index][n]
})
})
console.log(newArr)

Count all values in object where X is the first key letter

I would like to count all values where a letter appears first and return the letter with atleast half of all values in my object so for example I assuming I have an object like this
const sample = { "A,B,C": 4, "B,C,A": 3, "C,B,A": 2, "A,C,B": 2 };
I would return A because if you count all the values where A appears first you would get 6 (4+2)
This is what I currently have:
for (let votes of Object.values(sample)) {
sum += votes
}
stretchWin = Math.round(sum / 2)
winner = Object.entries(sample)
.filter(([, val]) => val >= stretchWin)
.map(([keys]) => keys)
With this I am getting an empty array because I am not counting all the values assigned to A

Iterate over the whole sample first to get a sum of the values by the first letter first, then iterate over that new object to identify which values match the target of half the total.
const sample = {
"A,B,C": 4,
"B,C,A": 3,
"C,B,A": 2,
"A,C,B": 2
};
const sumByChar = {};
for (const [key, value] of Object.entries(sample)) {
const char = key[0];
sumByChar[char] = (sumByChar[char] ?? 0) + value;
}
let sum = 0;
for (let votes of Object.values(sample)) {
sum += votes
}
const targetSum = Math.round(sum / 2);
const winners = Object.entries(sumByChar)
.filter(([, val]) => val >= targetSum)
.map(([key]) => key);
console.log(winners);

I'm not completely sure what you mean what the outcome should be. If I understand correctly you want something like this??
const sample = { "A,B,C": 4, "B,C,A": 3, "C,B,A": 2, "A,C,B": 2 };
const totalSum = Object.values(sample).reduce(
(previousValue, currentValue) => previousValue + currentValue,
0
);
const stretchWin = Math.round(totalSum / 2);
const winner = Object.entries(sample)
.filter(([key, value]) => {
const isFirstLetterA = key.startsWith("A");
return isFirstLetterA || value >= stretchWin;
})
.map(([key, value]) => value)
.reduce((previousValue, currentValue) => previousValue + currentValue, 0);
console.log(winner);

How to flip every character with the character next to it. Javascript

Need help inside the for loop to flip each character with the character before it.
function flip(str) {
//split string
//iterate through split string
//return joined string
var splitt = str.split('');
for(var i = 0; i < splitt.length; i++){
//flip every character with one before it
}
}
var output = flip('otatl');
console.log(output); // -> 'total'

function split(str) {
let splitt = str.split('');
for (let i=0; i<splitt.length-1; i+=2) {
const temp = splitt[i];
splitt[i] = splitt[i+1];
splitt[i+1] = temp;
}
return splitt.join('');
}

You can use ES6 destructuring assignment.
function flip(str) {
//split string
//iterate through split string
//return joined string
let splitt = str.split('');
for (let i=0; i < splitt.length; i++){
//flip every character with one before it
if (i%2 == 1) {
[splitt[i-1], splitt[i]] = [splitt[i], splitt[i-1]];
}
}
return splitt.join('');
}
let output = flip('otatl');
console.log(output); // -> 'total'
You can combine this technique with gillyb's loop pattern to reduce the iterations as follows:
function flip(str) {
//split string
//iterate through split string
//return joined string
let splitt = str.split('');
for (let i=1; i < splitt.length; i+=2){
//flip every character with one before it
[splitt[i-1], splitt[i]] = [splitt[i], splitt[i-1]];
}
return splitt.join('');
}
let output = flip('otatl');
console.log(output); // -> 'total'

Can do something similar with regex and array manipulation
const flip = (stringToFlip) => stringToFlip
.split(/(.{2})/) // array of strings of 2 chars
.map((e) => e.split('') // convert each string piece to array
.reverse() // reverse array
.join('') // convert array piece back to string
)
.join(''); // combine all parts
const result = flip('otatl');
console.log("flip('otatl')");
console.log(result);

If we're not restricted to for loops, this is my (slightly too code golf-ish?) answer:
const flip = (str) =>
str
.split('')
.reduce((a, v, i) => (a[i + ((i % 2) * -2 + 1)] = v, a), [])
.join('');
console.log(flip('otatl'));
console.log(flip('lfpi'));

I'm sorry I am late to the party but you can use reduce().
let input = "vanjskojfdghpja";
let output = input
.split('')
.reduce(([o,p],c,i) => i%2?[o+c+p,'']:[o,c],['',''])
.join('');
console.log(output);
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