Develop Reference

Why function output doesn't return an expected array

This should be something simple, but I miss why this code outputs 'undefined', while I expect to get an array with numbers: [1, 2]. I tried to debug it in console step by step, but still don't understand why newArr doesn't return from the function. Could someone explain, please.
function filterList(arr) {
let newArr = []
for(let i = 0; i <= arr.length; i++) {
if (typeof arr[i] !== "number") {
return
}
if (typeof arr[i] === 'number') {
newArr.push(arr[i])
}
}
return newArr
}
console.log(filterList([1,2,'a','b']))

The first return that is reached inside a function will end the function call and will return whatever you set there.
In your case, you are pushing 2 values in the array, but as soon as you reach the third one a you are just using return. This means that you are returning undefined. You should return newArr
function filterList(arr) {
let newArr = []
for(let i = 0; i <= arr.length; i++) {
if (typeof arr[i] !== "number") {
return newArr
}
if (typeof arr[i] === 'number') {
newArr.push(arr[i])
}
}
return newArr
}
console.log(filterList([1,2,'a','b']))

You're returning from the function immediately on finding an element that isn't a number, and the return value from the function is undefined not the array you've been patiently pushing numbers into. Just remove that statement altogether.
function filterList(arr) {
const newArr = [];
for (let i = 0; i <= arr.length; i++) {
if (typeof arr[i] === 'number') {
newArr.push(arr[i]);
}
}
return newArr;
}
console.log(filterList([1, 2, 'a', 'b', 3, 4]))

remove return in for loop
function filterList(arr) {
let newArr = []
for(let i = 0; i < arr.length; i++) {
if (typeof arr[i] === 'number') {
newArr.push(arr[i])
}
}
return newArr
}
const arr = [1,2,'a','b']
console.log(filterList(arr))
// with Array.filter
let result1 = arr.filter(ele => typeof ele === 'number')
console.log(result1)
// with Array.reduce
let result2 = arr.reduce((res, ele) => typeof ele === 'number' ? [...res, ele] : res, [])
console.log(result2)
function stopFunc(isStop = false) {
let total = 0
for(let i = 0; i < 10; i++) {
total += i
if(isStop) return 999
}
return total // return 45
}
console.log(stopFunc())
console.log(stopFunc(true))
you can check stopFunc function in my example, The return statement ends function execution and specifies a value to be returned to the function caller. . in your function :
if (typeof arr[i] !== "number") {
return
}
return omitted, undefined is returned instead.
You can check document return

Returning first n elements of an array Javascript

I am writing a function that will return the first n elements of an array and if n is undefined then return the first element.
*Edit: I solved the problem.
_.first = function(array, n) {
var result = [];
if (n == undefined) {
return array[0];
}
var m;
if (array.length < n) {
m = array.length;
} else {
m = n;
}
for (var i = 0; i < m; i++) {
result.push(array[i]);
} return result;
};

This program is basically checking if the n value is bigger than the array's length, if it is, then it exits.
If n is not a number it exits. If it is it executes the program and logs the values of indexes with for loop until i reaches the n value. Also, it pushes the values to the empty array, so you can get the values from the array for later use.
var arr1 = [2, 3, 4, 5, 6, 7, 8];
var arr2 = []; //empty array
function arrNreturn(arr, n){
if(typeof n != 'number'){
console.log('n is not a number');
return false; //exit the program
}
if(n > arr.length){
console.log("the n value is bigger than the length");
}else{
for(var i = 0; i < n; i++){
console.log(arr[n]);
arr2.push(arr[n]);
}
}
}
arrNreturn(arr1, 10);

1_.first = function(array, n) {
2 var result = [];
3 for (var i = 0; i < n; i++) {
4 result.push(array[i]);
5 return result;
6 } return array[0];
7};
the problem is on the 5th line, you seem to be returning the first and not taking into account the n parts
a solution might be
_first = function(array, n){
var result = [];
if(n === undefined) return array[0];
for (var i = 0; i < n; i++) {
result.push(array[i]);
}
return result;
}

Basic is
const newArr = function(arr) {return n }
Shorthand
const newArr = arr => { return n}

Why does my for loop only appear to run once?

This is one of freecodecamps challenges it passes the for loop inside the filter passes the first element of the array newArg but doesn't for the second one and so on therefore the challenge doesn't pass can someone explain to me why. Please don't write any full solutions as i just want a little help to move forward.
function destroyer(arr) {
// Remove all the values
var newArg = [];
for (var i=1; i < arguments.length; i++){
newArg.push(arguments[i]);
}
var newArray = arr.filter(function(val){
for (i = 0; i < newArg.length; i++) {
return val !== newArg[i];
}
});
return newArray;
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);

The filter call:
var newArray = arr.filter(function(val){
for (i = 0; i < newArg.length; i++) {
return val !== newArg[i];
}
});
is the same as:
var newArray = arr.filter(function(val) {
return val !== newArg[0];
});
because you are returning from the very first iteration.
Solution:
You'll have to wrap the return statement in an if like this:
var newArray = arr.filter(function(val){
for (i = 0; i < newArg.length; i++) {
if(val === newArg[i]) { // don't return if val !== newArg[i]
return true; // return only when they're ===
}
}
return false; // default return (nothing is found in the array)
});
Or use an alternative such as Array.prototype.some like this:
var newArray = arr.filter(function(val){
return newArg.some(function(arg) { // return true if some item pass the test (false otherwise)
return arg === val; // the test
})
});

Write an algorithm that takes an array and moves all of the zeros to the end

Write an algorithm that takes an array and moves all of the zeros to the end, preserving the order of the other elements.
For example:
moveZeros([false,1,0,1,2,0,1,3,"a"]) // returns[false,1,1,2,1,3,"a",0,0]
My code:
var moveZeros = function (arr) {
var zeros = [];
var others = [];
var res;
var arrayLength = arr.length;
for (var i = 0; i < arrayLength; i++) {
if (arr[i] == 0) {
zeros.push(arr[i]);
} else {
others.push(arr[i]);
}
}
var res = others.concat( zeros );
return res;
}
I get the following result:
Expected: ["a","b",null,"c","d",1,false,1,3,[],1,9,{},9,0,0,0,0,0,0,0,0,0,0],
Instead got: ["a","b",null,"c","d",1,1,3,1,9,{},9,0,0,0,false,0,0,[],0,0,0,0,0]
The expected result is quite close to what I achieved (see above) . I don't understand why I have false in a different place?

try this simply
var arr = [false,1,0,1,2,0,1,3,"a"];
arr.sort(function(a,b){if (a===0){return 1}});
document.body.innerHTML += JSON.stringify(arr);

Try to use a normal for loop and splice at this context to make your job done,
var arr = [false,1,0,1,2,0,1,3,"a"];
for(var i=arr.length;i>0;i--){
if(arr[i] === 0){ arr.push(arr.splice(i,1).pop()); }
}
console.log(arr); //[false, 1, 1, 2, 1, 3, "a", 0, 0]
document.body.innerHTML += JSON.stringify(arr);

Use Array#splice() and Array#push().
function moveZeros(a) {
var i = a.length - 1;
while (i--) {
if (a[i] === 0) {
a.push(a.splice(i, 1)[0]);
}
}
return a;
};
var array = [false, 1, 0, 1, 2, 0, 1, 3, "a"];
document.write('<pre>' + JSON.stringify(moveZeros(array), 0, 4) + '</pre>');
Another approach with swapping items.
function moveZeros(array) {
var length = array.length,
i = length;
while (--i) {
if (array[i] !== 0) continue;
length--;
while (i < length) {
[array[i + 1], array[i]] = [array[i], array[i + 1]];
i++;
}
}
return array;
}
var array = [false, 1, 0, 1, 2, 0, 1, 3, "a"];
console.log(...moveZeros(array));

Please use === operator to compare if value is 0:
if (arr[i] === 0) {
zeros.push(arr[i]);
} else {
others.push(arr[i]);
}

The code snippet below should work , please try , reason for null is that the length of the arr starts count from zero so when using <= you should subtract 1 from the total length.
var moveZeros = function (arr) {
// TODO: Program me
let zero = []
let others = []
let together = []
for (let i =0; i <= arr.length-1; i++){
if (arr[i] === 0){
zero.push(arr[i])
}
else{
others.push(arr[i])
}
}
together = others.concat(zero)
return together
}

var titleCase = function(title) {
var arr = [];
for (i = 0; i < title.length ; i++) {
if (title[i] !== 0 ) {
arr.push(title[i]);
}
}
for (i = 0 ; i < title.length ; i++) {
if (title[i] == 0 ) {
arr.push(title[i]);
}
}
return arr;
}

You could use the filter() method to achieve your goal. With the arrow notation you can shorten the code. For example: arr => arr === 0 is the shorthand for the anonymous filter function function(arr) { return arr === 0; }
var moveZeros = function (arr) {
const zeros = arr.filter (arr => arr === 0);
const others = arr.filter (arr => arr !== 0);
return others.concat(zeros);
}

var arr = [false,1,0,1,2,0,1,3,"a"]
function moveZeros(arr){
var zeros = [];
var others = [];
var output;
for (var i=0; i< arr.length; i++){
if (arr[i]===0){
zeros.push(arr[i]);
}else{
others.push(arr[i])
}
}
output = others.concat(zeros);
console.log(output);
}
moveZeros([false,1,0,1,2,0,1,3,"a"]);

return the first non repeating character in a string in javascript

So I tried looking for this in the search but the closest I could come is a similar answer in several different languages, I would like to use Javascript to do it.
The problem is I have an arbitrary string that I would like to return the first non repeating character. EX: 'aba' -> would return b
'aabcbd' -> would return c.
This is what I have so far, just a simple for loop to start.
var someString = 'aabcbd';
var firstNonRepeatedCharacter = function(string) {
for(var i = 0; i < someString.length; i++){
}
};
http://jsfiddle.net/w7F87/
Not sure where to go from here

You can use the indexOf method to find the non repeating character. If you look for the character in the string, it will be the first one found, and you won't find another after it:
function firstNonRepeatedCharacter(string) {
for (var i = 0; i < string.length; i++) {
var c = string.charAt(i);
if (string.indexOf(c) == i && string.indexOf(c, i + 1) == -1) {
return c;
}
}
return null;
}
Demo: http://jsfiddle.net/Guffa/Se4dD/

If you're looking for the first occurrence of a letter that only occurs once, I would use another data structure to keep track of how many times each letter has been seen. This would let you do it with an O(n) rather than an O(n2) solution, except that n in this case is the larger of the difference between the smallest and largest character code or the length of the string and so not directly comparable.
Note: an earlier version of this used for-in - which in practice turns out to be incredibly slow. I've updated it to use the character codes as indexes to keep the look up as fast as possible. What we really need is a hash table but given the small values of N and the small, relative speed up, it's probably not worth it for this problem. In fact, you should prefer #Guffa's solution. I'm including mine only because I ended up learning a lot from it.
function firstNonRepeatedCharacter(string) {
var counts = {};
var i, minCode = 999999, maxCode = -1;
for (i = 0; i < string.length; ++i) {
var letter = string.charAt(i);
var letterCode = string.charCodeAt(i);
if (letterCode < minCode) {
minCode = letterCode;
}
if (letterCode > maxCode) {
maxCode = letterCode;
}
var count = counts[letterCode];
if (count) {
count.count = count.count + 1;
}
else {
counts[letterCode] = { letter: letter, count: 1, index: i };
}
}
var smallestIndex = string.length;
for (i = minCode; i <= maxCode; ++i) {
var count = counts[i];
if (count && count.count === 1 && count.index < smallestIndex) {
smallestIndex = count.index;
}
}
return smallestIndex < string.length ? string.charAt(smallestIndex) : '';
}
See fiddle at http://jsfiddle.net/b2dE4/
Also a (slightly different than the comments) performance test at http://jsperf.com/24793051/2

var firstNonRepeatedCharacter = function(string) {
var chars = string.split('');
for (var i = 0; i < string.length; i++) {
if (chars.filter(function(j) {
return j == string.charAt(i);
}).length == 1) return string.charAt(i);
}
};
So we create an array of all the characters, by splitting on anything.
Then, we loop through each character, and we filter the array we created, so we'll get an array of only those characters. If the length is ever 1, we know we have a non-repeated character.
Fiddle: http://jsfiddle.net/2FpZF/

Two further possibilities, using ECMA5 array methods. Will return undefined if none exist.
Javascript
function firstNonRepeatedCharacter(string) {
return string.split('').filter(function (character, index, obj) {
return obj.indexOf(character) === obj.lastIndexOf(character);
}).shift();
}
console.log(firstNonRepeatedCharacter('aabcbd'));
On jsFiddle
Or if you want a bit better performance, especially on longer strings.
Javascript
function firstNonRepeatedCharacter(string) {
var first;
string.split('').some(function (character, index, obj) {
if(obj.indexOf(character) === obj.lastIndexOf(character)) {
first = character;
return true;
}
return false;
});
return first;
}
console.log(firstNonRepeatedCharacter('aabcbd'));
On jsFiddle

I came accross this while facing similar problem. Let me add my 2 lines.
What I did is a similar to the Guffa's answer. But using both indexOf method and lastIndexOf.
My mehod:
function nonRepeated(str) {
for(let i = 0; i < str.length; i++) {
let j = str.charAt(i)
if (str.indexOf(j) == str.lastIndexOf(j)) {
return j;
}
}
return null;
}
nonRepeated("aabcbd"); //c
Simply, indexOf() gets first occurrence of a character & lastIndexOf() gets the last occurrence. So when the first occurrence is also == the last occurence, it means there's just one the character.

Here's a Solution using Regex to replace all repeating characters and then returning the first character.
function firstNonRepeat(str) {
// Sorting str so that all repeating characters will come together & replacing it with empty string and taking first character using substr.
var rsl = str.split('').sort().join('').replace(/(\w)\1+/g,'').substr(0,1);
if(rsl) return rsl;
else return 'All characters are repeated in ' + str;
}
console.log(firstNonRepeat('aaabcccdeeef'));
console.log(firstNonRepeat('aaacbdcee'));
console.log(firstNonRepeat('aabcbd'));

First of all, start your loop at 1, not 0. There is no point in checking the first character to see if its repeating, obviously it can't be.
Now, within your loop, you have someString[i] and someString[i - 1]. They are the current and previous characters.
if someString[i] === someString[i - 1] then the characters are repeating, if someString[i] !== someString[i - 1] then they are not repeating, so you return someString[i]
I won't write the whole thing out for you, but hopefully the thought process behind this will help

function FirstNotRepeatedChar(str) {
var arr = str.split('');
var result = '';
var ctr = 0;
for (var x = 0; x < arr.length; x++) {
ctr = 0;
for (var y = 0; y < arr.length; y++) {
if (arr[x] === arr[y]) {
ctr+= 1;
}
}
if (ctr < 2) {
result = arr[x];
break;
}
}
return result;
}
console.log(FirstNotRepeatedChar('asif shaik'));

Here's an O(n) solution with 2 ES6 Sets, one tracking all characters that have appeared and one tracking only chars that have appeared once. This solution takes advantage of the insertion order preserved by Set.
const firstNonRepeating = str => {
const set = new Set();
const finalSet = new Set();
str.split('').forEach(char => {
if (set.has(char)) finalSet.delete(char);
else {
set.add(char);
finalSet.add(char);
}
})
const iter = finalSet.values();
return iter.next().value;
}

let arr = [10, 5, 3, 4, 3, 5, 6];
outer:for(let i=0;i<arr.length;i++){
for(let j=0;j<arr.length;j++){
if(arr[i]===arr[j+1]){
console.log(arr[i]);
break outer;
}
}
}
//or else you may try this way...
function firstDuplicate(arr) {
let findFirst = new Set()
for (element of arr)
if (findFirst.has(element ))
return element
else
findFirst.add(element )
}

function firstUniqChar(str) {
let myMap = new Map();
for(let i = 0; i < str.length; i++) {
let char = str.charAt(i);
if(!myMap.has(char)) {
myMap.set(char, 0);
}
myMap.set(char, myMap.get(char) + 1 );
}
for(let [key, value] of myMap) {
if(value === 1) {
return key;
}
}
return null;
}
let result = firstUniqChar("caabbdccee");
console.log(result);
You can use Map Object and set key and value, where in value you store the count for that particular character, After that you can iterate over map and check where is value 1 and return that key.
Map Object remembers the original insertion order of the keys.

This solution should works with array with integers and string.
function firstNoneRepeating(list, map = new Map()) {
for (let item of list) {
if (map.has(item)) {
map.set(item, map.get(item) + 1);
} else {
map.set(item, 1);
}
}
for (let [key, value] of map.entries()) {
if (value === 1) {
return key;
}
}
}
console.log(firstNoneRepeating("aabcbd"));
console.log(firstNoneRepeating([5, 2, 3, 4, 2, 6, 7, 1, 2, 3]));

let str='aabcbd'
let ans=''
for (let i=0;i<str.length;i++){
if(str.indexOf(str.charAt(i))===str.lastIndexOf(str.charAt(i))){
ans+=str.charAt(i)
break
}
}
console.log(ans)

Fill an empty array with zeros, with same length as the string array, and tally up how many times they appear through the loop. Grab the first one in the tallied array with a value of 1.
function firstNotRepeatingCharacter(s) {
const array = s.split("");
let scores = new Array(array.length).fill(0);
for (let char of array) {
scores[array.indexOf(char)]++;
}
const singleChar = array[scores.indexOf(1)];
return singleChar ? singleChar : "_"
}

You can iterate through each character to find() the first letter that returns a single match(). This will result in the first non-repeated character in the given string:
const first_nonrepeated_character = string => [...string].find(e => string.match(new RegExp(e, 'g')).length === 1);
const string = 'aabcbd';
console.log(first_nonrepeated_character(string)); // c

Here is my solution which have time complexity of o(n)
function getfirstNonRepeatingCharacterInAString(params) {
let count = {};
for (let i = 0; i < params.length; i++) {
let count1 = 0;
if (!count[params.charAt(i)]) {
count[params.charAt(i)] = count1 + 1;
}
else {
count[params.charAt(i)] = count[params.charAt(i)] + 1;
}
}
for (let key in count) {
if (count[key] === 1) {
return key;
}
}
return null;
}
console.log(getfirstNonRepeatingCharacterInAString("GeeksfoGeeks"));

Here is my solution using forEach and convert the string into an array
function firstNotRepeatingCharacter(s) {
var strArr = s.split("");
var found = "_";
strArr.forEach(function(item, index) {
if (strArr.indexOf(item) == index && strArr.indexOf(item, index + 1) == -1) {
if (found === "_") {
found = item;
}
}
})
return found;
}
firstNotRepeatingCharacter("abacabad")

Here is another approach:
Everytime you find equal chars store it in an array and break out of the loop. If the char is not found in the array then you have your first nonRepeating char
function nonRepeatingChars(value) {
const memory = []
for (let i = 0; i < value.length; i++) {
for (let j = i + 1; j < value.length; j++) {
if (value[i] === value[j]) {
memory.push(value[j])
break;
}
}
if (!memory.some(x => x === value[i])) {
return value[i];
}
}
return "all chars have duplicates";
}
console.log('First non repeating char is:',nonRepeatingChars("esen"))
console.log('First non repeating char is:',nonRepeatingChars("esesn"))
console.log('First non repeating char is:',nonRepeatingChars("eseulsn"))
console.log('First non repeating char is:',nonRepeatingChars("esesnn"))

> var firstNonRepeatedCharacter = function (str){
> for(i=0;i<str.length;i++){
> if(str.indexOf(str.charAt(i)) === str.lastIndexOf(str.charAt(i))){
> console.log(str.charAt(i));
> break;
> } } }
>
> firstNonRepeatedCharacter ("areerak");
you can check below link
https://codepen.io/t3veni/pen/wvvxJzm

Easy way to solve this algorithm, very straight forward.
function firstNonRepeatChar(str){
let map = {};
for(let i=0; i<str.length; i++){
if(Object.keys(map).includes(str[i])){
map[str[i]]++
}
else{
map[str[i]] = 1;
}
}
for(let j=0; j< Object.values(map).length; j++){
if(Object.values(map)[j] == 1){
console.log(Object.keys(map)[j]);
return
}
if (j == Object.values(map).length-1 && Object.values(map)[j] != 1){
console.log('_');
return;
}
else{
continue;
}
}
}
nonRepeat("aaabbcccdeeef");

Here is one other solution just using array, using 26 unique character as length of array:
var firstUniqChar = (function(s) {
var arr = [];
var str = s.toLowerCase();
for(let c of str){
let index = c.charCodeAt(0) - "a".charCodeAt(0);
arr[index]? ++arr[index]: arr[index]=1;
}
for(let c of str){
let index = c.charCodeAt(0) - 97;
if(arr[index] == 1){
return c;
};
}
return -1;
}("aabcbd"));
console.log(firstUniqChar);

We can keep track of frequency of each character of the string in an object.
For example : for "aabcbd" we can store the frequency as
{ "a":2, "b":2, "c":1, "d":1 }
This will take O(n) time.
Then we can traverse over this object and find the first character with frequency 1, which will also take O(n) time. So, the time complexity for this approach will be O(n).
const firstNonRepeating=(str)=>{
const obj={};
str.split("").forEach(item=>{
obj[item]
? obj[item]++
: obj[item]=1;
});
const item = Object.keys(obj).find(key=> obj[key] === 1);
return item;
}
Note: I use ES6 Object.keys method which may not work in older
browsers.

//To find first non repeating letter
//It will check for both upper and lower case
//only use one String.indexOf()
var mystr="ohvhvtccggt";
var checkFirstNonRepeating=function(){
var ele=[];
for(var i=0;i<mystr.length;i++) {
var key=mystr.charAt(i);
if(!ele[key])
ele[key]=0;
ele[key]++;
//Just check for second occurance of character
//no need to use indexOf twice
if(mystr.indexOf(key,i+1)==-1 && ele[key]<2)
return mystr[i];
}
return "All repeating letters";
}
console.log(checkFirstNonRepeating());
/*
Input : "ohvhvtoccggt"
Output : All repeating letters
Input :"oohjtht"
Output :j
*/

I used object to keep track of characters count in a string then return the char that has the fa value of 1. Here is a demo:
function firstNotRepeatingCharacter(s) {
// initialize an empty object to store chars
let seen = {};
let letter = '';
// iterate over each char in a string
// if it is already there increase value by one
// else set the value to 1
for(let char of s){
if (seen[char]){
seen[char] +=1;
} else {
seen[char] = 1;
}
}
// iterate over the new constructed object
// if the value is 1 and the output variable is empty
// return the associated key to the value 1
// else return '_'
for(let v in seen){
while(seen[v] == 1 && letter === ''){
letter += v;
return letter;
}
}
return('_');
}
console.log(firstNotRepeatingCharacter("abacabad"));
console.log(firstNotRepeatingCharacter("cbc"));
console.log(firstNotRepeatingCharacter("bcccccccccccccyb"));
console.log(firstNotRepeatingCharacter("aaa"));

The most satisfactory and easy to understand answer is the following.
function firstNotRepeatingCharacter(s) {
const arr = s.split("");
for(let i = 0; i < arr.length; i++){
let chr = arr[i];
if( arr.indexOf(arr[i]) == arr.lastIndexOf(arr[i])){
return arr[i]
}
}
return "_"
}
Explanation: It loops through all the characters of a string from forward and backward and then compares the values. If the index of both forward and backward search is true then it returns that character.

let str = 'aabbcdd';
let val = str.split('').reduce((a, e)=>{ if(str.indexOf(e) == str.lastIndexOf(e)) {a = e }; return a})
console.log(val); // c

the implementation below has a good time complexity and it accounts for letters with different cases:
steps
must touch every character in the string to know if it's repeated or not
function firstNonRepeatingLetter(wordd) {
const chars = {}
let word = wordd.toLowerCase()
// go through chars
// store chars in hash with values of array storing index of char and true if only 1 encountered so far
for (let i = 0; i < word.length; i += 1) {
let char = word[i]
if (chars[char]) {
chars[char][0] = false
} else {
chars[char] = [true, i]
}
}
let output = ''
let index;
for (let key in chars) {
// return char with true and lowest index
if (chars[key][0]) {
index = index === undefined ? chars[key][1] : index
if (index >= chars[key][1]) {
output = key
}
}
}
return index === undefined ? '' : wordd[index]
}
console.log(firstNonRepeatingLetter('sTreSS')) //T```

The bellow solution is a kind of frequency counter pattern and it will run only one loop, so O(n) will be the time complexity.
function firstNotRepeatingCharacter(str) {
const obj = {};
for (let i = 0, L = str.length; i < L; i++) {
const char = str[i];
obj[char] = obj[char] ? obj[char] + 1 : 1;
}
for (let key of Object.keys(obj)) {
if (obj[key] == 1) {
return key;
}
}
return -1;
}

Here is another solution
function firstNotRepeatingCharacter(s) {
const obj = {};
for (let i of s) {
if(!obj[i]) {
obj[i] = 1;
} else if (obj[i]) {
obj[i] = +obj[i] + 1;
}
}
for (let [key, value] of Object.entries(obj)) {
if(value == 1) return key;
}
return "_"
}

Using below method can achieve first non repeated character
function main(str) {
str = String(str).toLowerCase();
let non_repeated_char = 'N/A';
for (let i = 0; i < str.length; i++) {
let currentChar = str[i];
let repeated_times = String(str).split('').filter(e => e == currentChar).length;
if (repeated_times === 1) {
non_repeated_char = currentChar;
break;
}
}
return non_repeated_char;
};
let Result = main("basketball");
console.log("The Non Repeated char is-->", Result);