Arrange an array containing duplicate elements by prioritizing the first instances

Question

Arrange an array containing duplicate elements by prioritizing the first instances

I have a unique array of items that are duplicated, occurring twice right next to each other, such as:

const arr = ['1', '1', '2', '2', '3', '3']

My goal is to rearrange the array so that all unique values come first followed by their duplicates, like this:

const arr = ['1', '2', '3', '1', '2', '3']

Is there a way to achieve this using the sort() function or should I explore other methods?

I attempted the following approach but it did not yield the desired results:

const sorted = myDublicatedArray.sort((left, right) => left.index - right.index);

javascript arrays sorting

Answer 1

Answer №1

When tackling this problem using Array.sort() alone, it can present challenges. However, the following approach should be effective for handling both numeric and string inputs, unless specific requirements dictate otherwise:

const arrayWithDuplicates = ['1', '1', '2', '2', '3', '3'];
// convert to a Set to obtain unique values
const uniqueValues = new Set(arrayWithDuplicates);
// sort by value
const sortedArray = Array.from(uniqueValues).sort((a, b) => a - b);
// create a duplicate of the sorted array
const arrayOrganized = [...sortedArray].concat(sortedArray);

console.log(arrayOrganized)

Answer 2

When tackling this problem using Array.sort() alone, it can present challenges. However, the following approach should be effective for handling both numeric and string inputs, unless specific requirements dictate otherwise:

const arrayWithDuplicates = ['1', '1', '2', '2', '3', '3'];
// convert to a Set to obtain unique values
const uniqueValues = new Set(arrayWithDuplicates);
// sort by value
const sortedArray = Array.from(uniqueValues).sort((a, b) => a - b);
// create a duplicate of the sorted array
const arrayOrganized = [...sortedArray].concat(sortedArray);

console.log(arrayOrganized)

Answer 3

Answer №2

Based on the feedback received, it seems that there will be consecutive pairs of duplicate values in the input array. This means there will never be 3 or more duplicates in a row. To handle this scenario, you can utilize the following mapping technique:

const arrayWithDuplicates = ['1', '1', '2', '2', '3', '3'];
const result = arrayWithDuplicates.map((_, i, a) => a[i * 2 % a.length]);
console.log(result);

If speed is crucial for your implementation, consider using a traditional for loop approach:

const arr = ['1', '1', '2', '2', '3', '3'];
const result = [];
for (let j = 0; j < 2; j++)
    for (let i = 0; i < arr.length; i += 2)
        result.push(arr[i]);
console.log(result);

Answer 4

Based on the feedback received, it seems that there will be consecutive pairs of duplicate values in the input array. This means there will never be 3 or more duplicates in a row. To handle this scenario, you can utilize the following mapping technique:

const arrayWithDuplicates = ['1', '1', '2', '2', '3', '3'];
const result = arrayWithDuplicates.map((_, i, a) => a[i * 2 % a.length]);
console.log(result);

If speed is crucial for your implementation, consider using a traditional for loop approach:

const arr = ['1', '1', '2', '2', '3', '3'];
const result = [];
for (let j = 0; j < 2; j++)
    for (let i = 0; i < arr.length; i += 2)
        result.push(arr[i]);
console.log(result);

Answer 5

Answer №3

To achieve this, you can create an index for each value and then transform it into a grouped array followed by a flat array as the final output.

const
    array = ['1', '1', '2', '2', '3', '3'],
    result = array
        .reduce((indices => (r, v) => {
            indices[v] ??= 0;
            (r[indices[v]++] ??= []).push(v);
            return r;
        })({}), [])
        .flat();

console.log(result);

Answer 6

To achieve this, you can create an index for each value and then transform it into a grouped array followed by a flat array as the final output.

const
    array = ['1', '1', '2', '2', '3', '3'],
    result = array
        .reduce((indices => (r, v) => {
            indices[v] ??= 0;
            (r[indices[v]++] ??= []).push(v);
            return r;
        })({}), [])
        .flat();

console.log(result);

Answer 7

Answer №4

To achieve this, you can utilize the sort method in JavaScript. Begin by converting the elements into pairs of [element, occurrence]. For instance, the initial occurrence of 2 is represented as [2,0].

Subsequently, arrange the pairs first based on their occurrences, and in case of equal occurrences, sort them based on the element itself.

Lastly, extract only the element from each [element, occurrence] pair.

const arr = ['1', '1', '2', '2', '3', '3']

const result = Object.values(Object.groupBy(arr, e => e))
  .flatMap(arr => arr.map((e,i) => [e,i]))
  .sort(([a,i], [b,j]) => i-j || a-b)
  .map(([e]) => e)

console.log(result)

Answer 8

To achieve this, you can utilize the sort method in JavaScript. Begin by converting the elements into pairs of [element, occurrence]. For instance, the initial occurrence of 2 is represented as [2,0].

Subsequently, arrange the pairs first based on their occurrences, and in case of equal occurrences, sort them based on the element itself.

Lastly, extract only the element from each [element, occurrence] pair.

const arr = ['1', '1', '2', '2', '3', '3']

const result = Object.values(Object.groupBy(arr, e => e))
  .flatMap(arr => arr.map((e,i) => [e,i]))
  .sort(([a,i], [b,j]) => i-j || a-b)
  .map(([e]) => e)

console.log(result)

Answer 9

Answer №5

If the values are in ascending order and come in pairs, you can utilize the following method:

const arr = ['1', '1', '2', '2', '3', '3'];
const uniq = new Set(arr);
const desired = Array.from({ length: 2 }).flatMap(() => Array.from(uniq));

console.log(desired);

The code snippet above initializes a Set named uniq, containing all unique elements from the array. To achieve the desired outcome, we generate an array of length 2 using Array.from() and then expand it with flatMap(). By converting uniq from a Set to an array, we maintain the item sequence based on insertion order ('1'</, followed by '2'</, and finally '3'</).

In scenarios where the number of duplicates may vary dynamically (consistently across the entire array), you can enhance flexibility by incorporating

Array.from({ length: arr.length / uniq.size })

. This approach facilitates inputs like the ones shown below:

const arr = ['1', '2', '3'];
const arr = ['1', '1', '1', '2', '2', '2', '3', '3', '3'];

Note that the data elements should still be sorted and exhibit uniformity in duplicate quantity.

Answer 10

If the values are in ascending order and come in pairs, you can utilize the following method:

const arr = ['1', '1', '2', '2', '3', '3'];
const uniq = new Set(arr);
const desired = Array.from({ length: 2 }).flatMap(() => Array.from(uniq));

console.log(desired);

The code snippet above initializes a Set named uniq, containing all unique elements from the array. To achieve the desired outcome, we generate an array of length 2 using Array.from() and then expand it with flatMap(). By converting uniq from a Set to an array, we maintain the item sequence based on insertion order ('1'</, followed by '2'</, and finally '3'</).

In scenarios where the number of duplicates may vary dynamically (consistently across the entire array), you can enhance flexibility by incorporating

Array.from({ length: arr.length / uniq.size })

. This approach facilitates inputs like the ones shown below:

const arr = ['1', '2', '3'];
const arr = ['1', '1', '1', '2', '2', '2', '3', '3', '3'];

Note that the data elements should still be sorted and exhibit uniformity in duplicate quantity.

Arrange an array containing duplicate elements by prioritizing the first instances

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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