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let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);