Return a nested array

Question

I need help making this function after taking an array and another array (duplicate) that has just the numbers that are duplicated in the first array (for example array=[1,2,3,1,2,3,4,5,6,6], duplicate=[1,2,3,6]).

I want it to return an array as follow: finalArray1=[[1,1],[2,2],[3,3],4,5,[6,6]].

let input = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];

let sortArray = array => {
  return array.sort(function(a, b) {
    return a - b;});
}

function findDuplicates(data) {

  let duplicate = [];

  data.forEach(function(element, index) {

    // Find if there is a duplicate or not
    if (data.indexOf(element, index + 1) > -1) {

      // Find if the element is already in the duplicate array or not
      if (duplicate.indexOf(element) === -1) {
        duplicate.push(element);
      }
    }
  });

  return duplicate; 
}

let newArray = (array, duplicate) => { 

  for( var i = 0; i < 3; i++ ){
    for( var j = 0; j < 15; j++ ){
      if( duplicate[i] == array[j] ){
        let finalArray = new array().push(array[j]);
      }
    }
    return finalArray;
  }
}

Shouldn't duplicate also have 6? And : you never call the function findDuplicates. What if original has a value that occurs 3 times? Will duplicates have unique values still? How will this affect the output? — trincot
– trincot, Commented Mar 16, 2020 at 18:46
yeah my bad i'll edit that. Use can use console.log to see results but it doesnt work. I have some issue with the last function — Ali
– Ali, Commented Mar 16, 2020 at 18:49
no i already do the ordered result and find the duplicates, I just need help with the last function to put them all in separate arrays within the same array — Ali
– Ali, Commented Mar 16, 2020 at 18:50

Nina Scholz · Accepted Answer · 2020-03-16 19:04:27Z

2

You could take a Map and return in original order.

The map takes the items in insertation order, which means all item are later in original order. If a key exists, it createsd an array with the value. Otherwise just the item form the array is taken as value for the map.

At the end take only the values from the map and create an array of it.

let input = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20],
    result = Array.from(input
        .reduce((m, v) => m.set(v, m.has(v) ? [].concat(m.get(v), v) : v), new Map)
        .values()
    );

console.log(result);

A more traditional approach

let input = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20],
    result = input
        .sort((a, b) => a -b)
        .reduce((r, v, i, a) => {
            if (a[i - 1] !== v && v !== a[i + 1]) r.push(v); // check if unique
            else if (a[i - 1] !== v) r.push([v]);            // check left element
            else r[r.length - 1].push(v);
            return r;
        }, []);

console.log(result);

edited Mar 16, 2020 at 19:04

answered Mar 16, 2020 at 18:50

Nina Scholz

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2 Comments

Ali Over a year ago

Wow thanks a lot!! lol i was working on this since last night and you did it in 5lines.

Ali Over a year ago

could you just explain what you did on the result please

norbitrial · Accepted Answer · 2020-03-16 18:56:54Z

0

I think the following can work for you:

const input = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];

const inOrder = input.sort((a, b) => a - b);
const unique = inOrder.reduce((a, c) => {
  const found = a.find(e => e.includes(c));      
  if (found) found.push(c);
  else a.push([c]);      
  return a;
}, []);
const result = unique.map(e => e.length > 1 ? e : e[0]);

console.log(result);

I hope this helps!

edited Mar 16, 2020 at 18:56

answered Mar 16, 2020 at 18:51

norbitrial

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2 Comments

Ali Over a year ago

It works, thanks a lot. I'll go over it and if i dont understand something and i'll let you know if you dont mind

norbitrial Over a year ago

@Ali Sure, happy to help!

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