Remove all of the duplicate numbers in an array of numbers

Question

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

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

I'm taking this as wanting the final result as either:

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

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},

arr2 = arr.filter(function (e) {
   if (elems[e] === undefined) {
       elems[e] = true;
    return true;
  }
  return false;
});
console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

Answer 1

Using Set and Array.from()

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

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

Alternate using regex

regex explanation here

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);

Alternate using objects

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));

Answer 2

Just use a simple array.filter one-liner:

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);

You could use another filter statement if you wanted the second result:

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);

Answer 3

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)

For the second part you can use reduce()

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)

Answer 4

You can use closure and Map

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))

Answer 5

You can create both arrays in One Go

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)

Answer 6

You can use Array.prototype.reduce() create a hash object where the keys are the numbers in the array and the values are going to be the the repeated occurrence of numbers in the arr array variable..

Then using Object.keys():

• Remove all duplicates Object.keys(hash)
• Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

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);

Answer 7

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

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)

Answer 8

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

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));

Answer 9

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);