Eliminate any duplicated numbers present within an array of numerical values

Question

Eliminate any duplicated numbers present within an array of numerical values

Recently, I came across a puzzling question for practice that left me intrigued by the two possible outcomes it could be seeking.

Naturally, I am eager to explore both solutions.

Consider an array given as:

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

I interpret the question as requiring the final result to be either of these two possibilities:

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

OR:

let finalResult = [1, 9, 10];

The distinction lies in one solution removing all duplicate numbers while retaining the unique ones, and the other focusing on isolating the non-duplicate numbers.

Which leads me to the desire to craft two functions, each catering to one of the above scenarios.

An individual shared the following code snippet resulting in my second solution:

let elems = {},

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

As for the function necessary for the first scenario (removing all duplicates), I find myself uncertain.

javascript arrays duplicates

Answer 1

Answer №1

Utilizing the power of Set combined with 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)));

An alternative method using regex

Explanation of regex usage can be found 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);

An alternative approach 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

Utilizing the power of Set combined with 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)));

An alternative method using regex

Explanation of regex usage can be found 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);

An alternative approach 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 3

Answer №2

For a straightforward solution, try using the array.filter method like this:

let numbers = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let uniqueNumbers = numbers.filter((element, index, array) => array.indexOf(element) === index).sort((a, b) => a - b);
console.log(uniqueNumbers);

If you need another filter for different results, consider implementing it with an additional filter statement:

let numbers = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let nonRepeatingNumbers = numbers.filter((element, index, array) => array.filter(num => num === element).length === 1).sort((a, b) => a - b);
console.log(nonRepeatingNumbers);

Answer 4

For a straightforward solution, try using the array.filter method like this:

let numbers = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let uniqueNumbers = numbers.filter((element, index, array) => array.indexOf(element) === index).sort((a, b) => a - b);
console.log(uniqueNumbers);

If you need another filter for different results, consider implementing it with an additional filter statement:

let numbers = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let nonRepeatingNumbers = numbers.filter((element, index, array) => array.filter(num => num === element).length === 1).sort((a, b) => a - b);
console.log(nonRepeatingNumbers);

Answer 5

Answer №3

To eliminate duplicates in the first part, utilize Set() and Spread Syntax.

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, employ 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 does not 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 6

To eliminate duplicates in the first part, utilize Set() and Spread Syntax.

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, employ 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 does not 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 7

Answer №4

To achieve this, utilize the power of closure in combination with Map.

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

const buildUniqueArray = arr => {
  const mapObj = arr.reduce((acc, element) => {
    acc.has(element) ? acc.set(element, true) : acc.set(element, false)
    return acc
  }, new Map())
  
  return function(hasDuplicates = true) {
    if(hasDuplicates) return [...mapObj.keys()]
    else return [...mapObj].filter(([key, value]) => !value).map(([k, v])=> k)
  }
}

const uniqueNumbers = buildUniqueArray(numbers)

console.log(uniqueNumbers())
console.log(uniqueNumbers(false))

Answer 8

To achieve this, utilize the power of closure in combination with Map.

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

const buildUniqueArray = arr => {
  const mapObj = arr.reduce((acc, element) => {
    acc.has(element) ? acc.set(element, true) : acc.set(element, false)
    return acc
  }, new Map())
  
  return function(hasDuplicates = true) {
    if(hasDuplicates) return [...mapObj.keys()]
    else return [...mapObj].filter(([key, value]) => !value).map(([k, v])=> k)
  }
}

const uniqueNumbers = buildUniqueArray(numbers)

console.log(uniqueNumbers())
console.log(uniqueNumbers(false))

Answer 9

Answer №5

You have the ability to generate two arrays simultaneously

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 10

You have the ability to generate two arrays simultaneously

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 11

Answer №6

To create a hash object in JavaScript, you can utilize the Array.prototype.reduce() method. This will assign numbers in an array as keys and their corresponding repeated occurrences as values in the hash object.

Afterwards, you can make use of the Object.keys() method:

Eliminate duplicates by accessing Object.keys(hash)
Filter out duplicates to obtain numbers with only one occurrence using Array.prototype.filter()

Here is the code snippet for illustration:

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), {});

// Resulting in: [1, 2, 3, 4, 5, 8, 9, 10]
const finalResultOne = Object.keys(hash);

// Output: [1, 9, 10]
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);

console.log('Final Result One:', ...finalResultOne);
console.log('Final Result Two:', ...finalResultTwo);

Answer 12

To create a hash object in JavaScript, you can utilize the Array.prototype.reduce() method. This will assign numbers in an array as keys and their corresponding repeated occurrences as values in the hash object.

Afterwards, you can make use of the Object.keys() method:

Eliminate duplicates by accessing Object.keys(hash)
Filter out duplicates to obtain numbers with only one occurrence using Array.prototype.filter()

Here is the code snippet for illustration:

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), {});

// Resulting in: [1, 2, 3, 4, 5, 8, 9, 10]
const finalResultOne = Object.keys(hash);

// Output: [1, 9, 10]
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);

console.log('Final Result One:', ...finalResultOne);
console.log('Final Result Two:', ...finalResultTwo);

Answer 13

Answer №7

To prevent duplicates in an array, you have the option to either sort the array first and then filter it by checking for duplicates on one side 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 14

To prevent duplicates in an array, you have the option to either sort the array first and then filter it by checking for duplicates on one side 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 15

Answer №8

Just like many others have pointed out, the solution for the first question is simply using [...new Set(arr)]

To solve the second question, you can filter out elements that occur more than once in the array:

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 16

Just like many others have pointed out, the solution for the first question is simply using [...new Set(arr)]

To solve the second question, you can filter out elements that occur more than once in the array:

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 17

Answer №9

Here is a code snippet that removes duplicates from an array and prints the unique elements:

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);

Answer 18

Here is a code snippet that removes duplicates from an array and prints the unique elements:

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);

Eliminate any duplicated numbers present within an array of numerical values

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

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Eliminate any duplicated numbers present within an array of numerical values

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

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