We Keep Coding

Two Arrays: Find items that are missing in the first, find items that are missing in the second

Imagine two arrays:
const array1 = [1,2,3];
const array2 = [2,3,4];
Now, I want to get all the differences of these two arrays and put them in two new arrays.
One Array will be for all the items that where missing in the first.
The other will be for the items missing in the second.
The result would look something like this:
const newArray1 = [1];
const newArray2 = [4];
How would I go about this and what is the most efficient way?

const array1 = [2,1,3,5,2,1,3,5];
const array2 = [4,3,2,6,7,4,3,2,6,7];
function diff(arr1, arr2) {
const dontAddDuplicates = true;
arr1.sort();
arr2.sort();
let a1 = [];
let a2 = [];
let i = 0;
let j = 0;
while (i < array1.length || j < array2.length) {
if (i >= arr1.length) {
if (!dontAddDuplicates || (a2.length == 0 || a2[a2.length - 1] != arr2[j])) {
a2.push(arr2[j]);
}
j++;
} else if (j >= array2.length) {
if (!dontAddDuplicates || (a1.length == 0 || a1[a1.length - 1] != arr1[i])) {
a1.push(arr1[i]);
}
i++;
} else if (arr1[i] < arr2[j]) {
if (!dontAddDuplicates || (a1.length == 0 || a1[a1.length - 1] != arr1[i])) {
a1.push(arr1[i]);
}
i++;
} else if (arr2[j] < arr1[i]) {
if (!dontAddDuplicates || (a2.length == 0 || a2[a2.length - 1] != arr2[j])) {
a2.push(arr2[j]);
}
j++;
} else {
// Same value, do nothing
i++;
j++;
}
}
return [a1, a2];
}
console.log(diff(array1, array2));
// OUTPUT: [[1, 5], [4, 6, 7]]
Here's another potential implementation using sorting, but it has the side effect of leaving array1 and array2 in a sorted fashion. Sorting allows you to avoid needing to rescan the other array every time. If they are already sorted then great you can skip this step. If the side effect is a problem, then use a deep copy of array1 and array2 before calling sort.
Flip dontAddDuplicates if you want duplicates or not. I notice the other implementations don't account for that, but easy enough to add.
Run time should be: SORT N + SORT M + N + M = SORT N = N LOG N depending on your input sizes and distributions SORT will be the significant O Notation https://www.bigocheatsheet.com/
https://jsfiddle.net/buscgtL2/1/
If you want to do it in N + M + N + M = N time you can use this implementation which uses a hash map instead of sorting. This has a disadvantage on memory space.
const array1 = [2,1,3,5,2,1,3,5];
const array2 = [4,3,2,6,7,4,3,2,6,7];
function diff(arr1, arr2) {
let dontAddDuplicates = true;
let a1 = [];
let a2 = [];
let a1hash = {};
let a2hash = {};
for (let i = 0; i < arr1.length; i++) {
a1hash[arr1[i]] = 0;
}
for (let i = 0; i < arr2.length; i++) {
a2hash[arr2[i]] = 0;
}
for (let i = 0; i < arr1.length; i++) {
if (!a2hash.hasOwnProperty(arr1[i])) {
if (!dontAddDuplicates || a1hash[arr1[i]] == 0) {
a1hash[arr1[i]] = 1;
a1.push(arr1[i]);
}
}
}
for (let i = 0; i < arr2.length; i++) {
if (!a1hash.hasOwnProperty(arr2[i])) {
if (!dontAddDuplicates || a2hash[arr2[i]] == 0) {
a2hash[arr2[i]] = 1;
a2.push(arr2[i]);
}
}
}
return [a1, a2];
}
console.log(diff(array1, array2));
//OUTPUT: [[1, 5], [4, 6, 7]]
https://jsfiddle.net/2945y3an/1/
The worse performance would be any algorithm where for every N element you scan array M searching for a match. This would be N * M = N^2

You can achieve it in very simple way with minimum line of codes by using Array.filter() along with Array.includes() methods of JavaScript.
Working Demo :
const array1 = [1,2,3];
const array2 = [2,3,4];
const updatedArray1 = array1.filter(item => !array2.includes(item));
const updatedArray2 = array2.filter(item => !array1.includes(item));
console.log(updatedArray1); // [1]
console.log(updatedArray2); // [4]

Something like this would be relatively efficient, you're only iterating through two arrays.
Iterate through the first array and add any missing items to the first newArray. Repeat the same process for the second array.
If you need to only loop through the array once, you would need to do something similar to what is done below, but only loop through the longest array.
const array1 = [1,2,3];
const array2 = [2,3];
function diff (arr1, arr2) {
const newA1 = [];
const newA2 = [];
arr1.forEach((item, i) => {
let index = arr2.findIndex(it => it === item);
if(index < 0) newA1.push(item);
});
arr2.forEach((item, i) => {
let index = arr1.findIndex(it => it === item);
if(index < 0) newA2.push(item);
});
return [newA1, newA2];
}
console.log(diff(array1, array2));
More efficent method (only loop through 1 array). This way you choose the longest array, loop through it, and check for duplicates of the current item of the long array and if an item in the secondary array exists at the same position, check for duplicates of this item in the longest array as well. This method is similar to the method above, but there is only 1 for each loop.
const array1 = [1,2,3];
const array2 = [3,4,5];
function diff (arr1, arr2) {
const newA1 = [];
const newA2 = [];
let baseArr, secondaryArr;
if(arr1.length > arr2.length) {
baseArr = arr1;
secondaryArr = arr2;
} else {
baseArr = arr2;
secondaryArr = arr1;
}
baseArr.forEach((item, i) => {
const secondaryArrI = secondaryArr.findIndex(it => it === item);
if(secondaryArrI < 0) newA1.push(item)
if(typeof secondaryArr[i] !== "undefined") {
const removeI = baseArr.findIndex(it => it === secondaryArr[i]);
if(removeI < 0) newA2.push(secondaryArr[i]);
}
})
return [newA1, newA2];
}
console.log(diff(array1, array2));

Lodash difference if you dot't mind
const array1 = [1,2,3];
const array2 = [2,3,4];
console.log(_.difference(array1, array2));
console.log(_.difference(array2, array1));
.as-console-wrapper{min-height: 100%!important; top: 0}
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.21/lodash.min.js"></script>

JS delete duplicated items from array without higher order functions

I know it's a stupid question, but I only learning programming 3 months now.
How would you solve this problem, if you can't use higher order functions and built-in methods, like filter or indexOf?
Create a function that takes a list of numbers and returns a new list where all the duplicate values are removed
I got this so far, but I think It's a dead end...
const array = [1, 2, 3, 3, 1];
const removeDuplicate = () => {
let shortArray = [];
let index = 0;
for (let i = 0; i < array.length; i++) {
for (let j = 0; j < array.length; j++) {
if (i != j) {
if (array[i] == array[j]) {
shortArray[index] += array[i]
console.log(array[i]);
}
}
}
}
return shortArray;
}
console.log(removeDuplicate());
return this:
1
3
3
1
[ NaN ]
thanks!

Use an object as a helper. If a value appears in the helper, it's not unique and can be ignored. If it's not in the helper it's unique, push it into the result array, and add it to the helper object.
const array = [1, 2, 3, 3, 1];
const removeDuplicate = (arr) => {
const helperMap = {};
const result = [];
for (let i = 0; i < arr.length; i++) {
const item = arr[i];
if (!helperMap[item]) {
result[result.length] = item;
helperMap[item] = true;
}
}
return result;
};
console.log(removeDuplicate(array));

function unique(arr) {
var obj = {};
for (var i = 0; i < arr.length; i++) {
var value = arr[i];
obj[value] = true; // set as key
}
return Object.keys(obj); //return all keys
}

Use below function:
function RemoveDuplicate(array){
let shortArray = [];
let index = 0;
for (let i = 0; i < array.length; i++) {
let exist=false;
for(let j=0;j<shortArray.length;j++){
if(array[i]==shortArray[j]){
exist=true;
break;
}
}
if(!exist){
shortArray[shortArray.length]=array[i];
}
}
return shortArray;
}

Get elements that has no double value in an array

I have a question . How do you retrieve elements that has no double value in an array?? For example: [1,1,2,2,3,4,4,5] then you retrieve [3,5] only.
Thanks in advance
for (var j = 0; j < newArr.length; j++) {
if ((arr1.indexOf(newArr[j]) === 0) && (arr2.indexOf(newArr[j]) === 0)) {
index = newArr.indexOf(j); newArr.splice(index, 1);
}
}

If the item in the array is unique then the index found from the beginning should equal the index found from the end, in other words:
var xs = [1, 1, 2, 2, 3, 4, 4, 5];
var result = xs.filter(function(x) {
return xs.indexOf(x) === xs.lastIndexOf(x);
});
console.log(result); //=> [3, 5]

sorry for the presentation its my first post !
You have to compare each element of your array to the others in order to get the number of occurence of each element
var tab = [1,1,2,2,3,4,4,5] //The array to analyze
tab = tab.sort(); // we sort the array
show(tab); // we display the array to the console (F12 to open it)
var uniqueElementTab = []; // this array will contain all single occurence
var sameElementCounter = 0;
for(x=0;x<tab.length;x++){ // for all element in the array
sameElementCounter = 0;
for(y=0;y<tab.length;y++){ // we compare it to the others
if((tab[x]==tab[y])){
sameElementCounter+=1; // +1 each time we meet the element elsewhere
}
}
if(sameElementCounter<=1){
uniqueElementTab.push(tab[x]); //if the element is unique we add it to a new array
}
}
show(uniqueElementTab); // display result
function show(tab) { // Simple function to display the content of an array
var st="";
for(i=0;i<tab.length;i++){
st += tab[i]+" ";
}
console.log(st+"\n");
}
Hope it helps.

Here is a simple "tricky" solution using Array.sort, Array.join, Array.map, String.replace and String.split functions:
var arr = [1, 1, 2, 2, 3, 4, 4, 5];
arr.sort();
var unique = arr.join("").replace(/(\d)\1+/g, "").split("").map(Number);
console.log(unique); // [3, 5]

create new array tmp,and check already value exist by indexOf .If existed delete by splice function..
var arr = [1,1,2,2,3,4,4,5];
var tmp = [];
var dup = [];
for(var i = 0; i < arr.length; i++){
var ind = tmp.indexOf(arr[i]);
if(ind == -1){
if(dup.indexOf(arr[i]) == -1){
tmp.push(arr[i]);
}
}
else{
tmp.splice(ind,1);
dup.push(arr[i]);
}
}
console.log(tmp);

This would be my way of doing this job.
var arr = [1,1,2,2,3,4,4,5],
uniques = Object.keys(arr.reduce((p,c) => (c in p ? Object.defineProperty(p, c, {enumerable : false,
writable : true,
configurable : true})
: p[c] = c,
p), {}));
console.log(uniques);

A solution for unsorted arrays with a hash table for the items. Complexity O(2n)
var array = [1, 1, 2, 2, 3, 4, 4, 5, 1],
hash = Object.create(null),
single;
array.forEach(function (a, i) {
hash[a] = a in hash ? -1 : i;
});
single = array.filter(function (a, i) {
return hash[a] === i;
});
console.log(single);

If the array is sorted, you can solve this in O(n) (see "pushUniqueSinglePass" below):
function pushUniqueSinglePass(array, unique) {
var prev; // last element seen
var run = 0; // number of times it has been seen
for (var i = 0; i < array.length; i++) {
if (array[i] != prev) {
if (run == 1) {
unique.push(prev); // "prev" appears only once
}
prev = array[i];
run = 1;
} else {
run++;
}
}
}
function pushUniqueWithSet(array, unique) {
var set = new Set();
for (var i = 0; i < array.length; i++) {
set.add(array[i]);
}
for (let e of set) {
unique.push(set);
}
}
// Utility and test functions
function randomSortedArray(n, max) {
var array = [];
for (var i = 0; i < n; i++) {
array.push(Math.floor(max * Math.random()));
}
return array.sort();
}
function runtest(i) {
var array = randomSortedArray(i, i / 2);
var r1 = [],
r2 = [];
console.log("Size: " + i);
console.log("Single-pass: " + time(
pushUniqueSinglePass, array, r1));
console.log("With set: " + time(
pushUniqueWithSet, array, r2));
// missing - assert r1 == r2
}
[10, 100, 1000, 10000,
100000, 1000000
].forEach(runtest);
function time(fun, array, unique) {
var start = new Date().getTime();
fun(array, unique);
return new Date().getTime() - start;
}
This is much more efficient than using maps or sorting (time it!). In my machine, a 1M sorted array can have its unique elements found in 18 ms; while the version that uses a set requires 10x more.

Check if an array contains duplicate values [duplicate]

This question already has answers here:
In Javascript, how do I check if an array has duplicate values?
(9 answers)
Closed 10 months ago.
I wanted to write a javascript function which checks if array contains duplicate values or not.
I have written the following code but its giving answer as "true" always.
Can anybody please tell me what am I missing.
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
}
return false; // means there are no duplicate values.
}

An easy solution, if you've got ES6, uses Set:
function checkIfArrayIsUnique(myArray) {
return myArray.length === new Set(myArray).size;
}
let uniqueArray = [1, 2, 3, 4, 5];
console.log(`${uniqueArray} is unique : ${checkIfArrayIsUnique(uniqueArray)}`);
let nonUniqueArray = [1, 1, 2, 3, 4, 5];
console.log(`${nonUniqueArray} is unique : ${checkIfArrayIsUnique(nonUniqueArray)}`);

let arr = [11,22,11,22];
let hasDuplicate = arr.some((val, i) => arr.indexOf(val) !== i);
// hasDuplicate = true
True -> array has duplicates
False -> uniqe array

This should work with only one loop:
function checkIfArrayIsUnique(arr) {
var map = {}, i, size;
for (i = 0, size = arr.length; i < size; i++){
if (map[arr[i]]){
return false;
}
map[arr[i]] = true;
}
return true;
}

You got the return values the wrong way round:
As soon as you find two values that are equal, you can conclude that the array is not unique and return false.
At the very end, after you've checked all the pairs, you can return true.
If you do this a lot, and the arrays are large, you might want to investigate the possibility of sorting the array and then only comparing adjacent elements. This will have better asymptotic complexity than your current method.

Assuming you're targeting browsers that aren't IE8,
this would work as well:
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
if (myArray.indexOf(myArray[i]) !== myArray.lastIndexOf(myArray[i])) {
return false;
}
}
return true; // this means not unique
}

Here's an O(n) solution:
function hasDupes(arr) {
/* temporary object */
var uniqOb = {};
/* create object attribute with name=value in array, this will not keep dupes*/
for (var i in arr)
uniqOb[arr[i]] = "";
/* if object's attributes match array, then no dupes! */
if (arr.length == Object.keys(uniqOb).length)
alert('NO dupes');
else
alert('HAS dupes');
}
var arr = ["1/1/2016", "1/1/2016", "2/1/2016"];
hasDupes(arr);
https://jsfiddle.net/7kkgy1j3/

Another solution:
Array.prototype.checkIfArrayIsUnique = function() {
this.sort();
for ( var i = 1; i < this.length; i++ ){
if(this[i-1] == this[i])
return false;
}
return true;
}

function hasNoDuplicates(arr) {
return arr.every(num => arr.indexOf(num) === arr.lastIndexOf(num));
}
hasNoDuplicates accepts an array and returns true if there are no duplicate values. If there are any duplicates, the function returns false.

Without a for loop, only using Map().
You can also return the duplicates.
(function(a){
let map = new Map();
a.forEach(e => {
if(map.has(e)) {
let count = map.get(e);
console.log(count)
map.set(e, count + 1);
} else {
map.set(e, 1);
}
});
let hasDup = false;
let dups = [];
map.forEach((value, key) => {
if(value > 1) {
hasDup = true;
dups.push(key);
}
});
console.log(dups);
return hasDup;
})([2,4,6,2,1,4]);

Late answer but can be helpful
function areThereDuplicates(args) {
let count = {};
for(let i = 0; i < args.length; i++){
count[args[i]] = 1 + (count[args[i]] || 0);
}
let found = Object.keys(count).filter(function(key) {
return count[key] > 1;
});
return found.length ? true : false;
}
areThereDuplicates([1,2,5]);

The code given in the question can be better written as follows
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = i+1; j < myArray.length; j++)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
return false; // means there are no duplicate values.
}

Returns the duplicate item in array and creates a new array with no duplicates:
var a = ["hello", "hi", "hi", "juice", "juice", "test"];
var b = ["ding", "dong", "hi", "juice", "juice", "test"];
var c = a.concat(b);
var dupClearArr = [];
function dupArray(arr) {
for (i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) != i && arr.indexOf(arr[i]) != -1) {
console.log('duplicate item ' + arr[i]);
} else {
dupClearArr.push(arr[i])
}
}
console.log('actual array \n' + arr + ' \nno duplicate items array \n' + dupClearArr)
}
dupArray(c);

const containsMatches = (a1, a2) => a1.some((v) => a2.includes(v));

If your array nests other arrays/objects, using the Set approach may not be what you want since comparing two objects compares their references. If you want to check that their contained values are equal, something else is needed. Here are a couple different approaches.
Approach 1: Map using JSON.stringify for keys
If you want to consider objects with the same contained values as equal, here's one simple way to do it using a Map object. It uses JSON.stringify to make a unique id for each element in the array.
I believe the runtime of this would be O(n * m) on arrays, assuming JSON.stringify serializes in linear time. n is the length of the outer array, m is size of the arrays. If the objects get very large, however, this may slow down since the keys will be very long. Not a very space-efficient implementation, but it is simple and works for many data types.
function checkArrayDupeFree(myArray, idFunc) {
const dupeMap = new Map();
for (const el of myArray) {
const id = idFunc(el);
if (dupeMap.has(id))
return false;
dupeMap.set(id, el);
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFree(notUnique, JSON.stringify)}`);
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFree(unique, JSON.stringify)}`);
Of course, you could also write your own id-generator function, though I'm not sure you can do much better than JSON.stringify.
Approach 2: Custom HashMap, Hashcode, and Equality implementations
If you have a lot of big arrays, it may be better performance-wise to implement your own hash/equality functions and use a Map as a HashMap.
In the following implementation, we hash the array. If there is a collision, map a key to an array of collided values, and check to see if any of the array values match according to the equality function.
The downside of this approach is that you may have to consider a wide range of types for which to make hashcode/equality functions, depending on what's in the array.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
Here's a demo of the custom HashMap in action. I implemented a hashing function and an equality function for arrays of arrays.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
function arrayHasher(arr) {
let hash = 19;
for (let i = 0; i < arr.length; i++) {
const el = arr[i];
const toHash = Array.isArray(el)
? arrayHasher(el)
: el * 23;
hash = hash * 31 + toHash;
}
return hash;
}
function arrayEq(a, b) {
if (a.length != b.length)
return false;
for (let i = 0; i < a.length; i++) {
if ((Array.isArray(a) || Array.isArray(b)) && !arrayEq(a[i], b[i]))
return false;
else if (a[i] !== b[i])
return false;
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFreeWHashes(notUnique, arrayHasher, arrayEq)}`);
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFreeWHashes(unique, arrayHasher, arrayEq)}`);

function checkIfArrayIsUnique(myArray)
{
isUnique=true
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
isUnique=false
}
}
}
}
return isUnique;
}
This assume that the array is unique at the start.
If find two equals values, then change to false

i think this is the simple way
$(document).ready(function() {
var arr = [1,2,3,9,6,5,6];
console.log( "result =>"+ if_duplicate_value (arr));
});
function if_duplicate_value (arr){
for(i=0;i<arr.length-1;i++){
for(j=i+1;j<arr.length;j++){
if(arr[i]==arr[j]){
return true;
}
}
}
return false;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

var c=[2,2,3,3,5,5,4,4,8,8];
for(var i=0; i<b.length; i++){
for(var j=i+1; j<b.length; j++){
if(c[i]==c[j]){
console.log(c[j]);
}
}
}

Remove duplicate item from array Javascript [duplicate]

This question already has answers here:
Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array
(97 answers)
Closed 8 years ago.
I'm looking for an easy way of removing a duplicate value from an array. I figured out how to detect if there is a duplicate or not, just I don't know how to "push" it from the value. For example, if you go to the link provided, and then type, "abca" (press return/enter key after each letter).. it will alert "duplicate!"
But I also want to figure out how to remove that duplicate from the textarea?
http://jsfiddle.net/P3gpp/
This is the part that seems to not be working ::
sort = sort.push(i);
textVal = sort;
return textVal;

Why do it the hard way, it can be done more easily using javascript filter function which is specifically for this kind of operations:
var arr = ["apple", "bannana", "orange", "apple", "orange"];
arr = arr.filter( function( item, index, inputArray ) {
return inputArray.indexOf(item) == index;
});
---------------------
Output: ["apple", "bannana", "orange"]

Based on user2668376 solution, this will return a new array without duplicates.
Array.prototype.removeDuplicates = function () {
return this.filter(function (item, index, self) {
return self.indexOf(item) == index;
});
};
After that you can do:
[1, 3, 3, 7].removeDuplicates();
Result will be; [1, 3, 7].

These are the functions I created/use for removing duplicates:
var removeDuplicatesInPlace = function (arr) {
var i, j, cur, found;
for (i = arr.length - 1; i >= 0; i--) {
cur = arr[i];
found = false;
for (j = i - 1; !found && j >= 0; j--) {
if (cur === arr[j]) {
if (i !== j) {
arr.splice(i, 1);
}
found = true;
}
}
}
return arr;
};
var removeDuplicatesGetCopy = function (arr) {
var ret, len, i, j, cur, found;
ret = [];
len = arr.length;
for (i = 0; i < len; i++) {
cur = arr[i];
found = false;
for (j = 0; !found && (j < len); j++) {
if (cur === arr[j]) {
if (i === j) {
ret.push(cur);
}
found = true;
}
}
}
return ret;
};
So using the first one, this is how your code could look:
function cleanUp() {
var text = document.getElementById("fld"),
textVal = text.value,
array;
textVal = textVal.replace(/\r/g, " ");
array = textVal.split(/\n/g);
text.value = removeDuplicatesInPlace(array).join("\n");
}
DEMO: http://jsfiddle.net/VrcN6/1/

You can use Array.reduce() to remove the duplicates. You need a helper object to keep track of how many times an item has been seen.
function cleanUp()
{
var textBox = document.getElementById("fld"),
array = textBox.value.split(/\r?\n/g),
o = {},
output;
output = array.reduce(function(prev, current) {
var key = '$' + current;
// have we seen this value before?
if (o[key] === void 0) {
prev.push(current);
o[key] = true;
}
return prev;
}, []);
// write back the result
textBox.value = output.join("\n");
}
The output of the reduce() step can be used directly to populate the text area again, without affecting the original sort order.
Demo

You can do this easily with just an object:
function removeDuplicates(text) {
var seen = {};
var result = '';
for (var i = 0; i < text.length; i++) {
var char = text.charAt(i);
if (char in seen) {
continue;
} else {
seen[char] = true;
result += char;
}
}
return result;
}
function cleanUp() {
var elem = document.getElementById("fld");
elem.value = removeDuplicates(elem.value);
}

arr3 = [1, 2, 3, 2, 4, 5];
unique = [];
function findUnique(val)
{
status = '0';
unique.forEach(function(itm){
if(itm==val){
status=1;
}
})
return status;
}
arr3.forEach(function(itm){
rtn = findUnique(itm);
if(rtn==0)
unique.push(itm);
});
console.log(unique); // [1, 2, 3, 4, 5]