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Javascript - Mix two arrays of objects ordering by date and without repetitions

Time:01-03

I have two lists:

let originalDocs = [
  {
    id: "sara",
    date: new Date("01/01/2022")
  },
  {
    id: "vic",
    date: new Date("10/26/2020")
  }
];

let newDocs = [
   {
     id: "vic",
     date: new Date("01/02/2022")
   },
   {
     id: "raul",
     date: new Date("05/05/2021")
   }
];

I need to mix both lists, ordering by date (in descending order) and without repetitions.

In order to do that, I have decided to:

  1. Delete the set of elements of the original array that are included in the new array (I mean, deleting the intersection) without modifying the original array.
  2. Mix the resulted array ordering by date.

This is my current code:

function removeIntersection(originalDocs, newDocs) {
  return  originalDocs.filter((originalDoc) => {
    const index = newDocs.findIndex(
      (newDoc) => newDoc.id === originalDoc.id
    );

    return index === -1;
  });
}

function mixInOrder(originalDocs, newDocs) {
   return [...newDocs, ...originalDocs]
    .sort((a, b) => b.date - a.date);
}


//
// MAIN
//

let originalDocs = [
  {
    id: "sara",
    date: new Date("01/01/2022")
  },
  {
    id: "vic",
    date: new Date("10/26/2020")
  }
];

let newDocs = [
  {
    id: "vic",
    date: new Date("01/02/2022")
  },
  {
    id: "raul",
    date: new Date("05/05/2021")
  }
];

const result = mixInOrder(
  removeIntersection(originalDocs, newDocs),
  newDocs
);

console.log(result);

How can I do the same in a more optimal way, I mean, without the need of merging (using the spread syntax) the list before sorting it, or deleting the intersection in O(1).

I mean, is it possible to just insert in order avoiding the copy?

CodePudding user response:

You can easily achieve the result using Set and reduce as:

  • Create a new array by adding the originalDocs first and then newDocs afterwards.
  • Use reduce to loop over the newly created array and then filter the objects.

You can use Set to check for existence of a key in efficient way

  • After filtering, you can then sort it accordingly.

let originalDocs = [
  {
    id: 'sara',
    date: new Date('01/01/2022'),
  },
  {
    id: 'vic',
    date: new Date('10/26/2020'),
  },
];

let newDocs = [
  {
    id: 'vic',
    date: new Date('01/02/2022'),
  },
  {
    id: 'raul',
    date: new Date('05/05/2021'),
  },
];

const set = new Set();
const result = [...originalDocs, ...newDocs].reduce((acc, curr) => {
    if (!set.has(curr.id)) {
      acc.push(curr);
      set.add(curr.id);
    }
    return acc;
  }, [])
  .sort((a, b) => b.date - a.date);
console.log(result);

CodePudding user response:

Here's how to do this using a JS Map, where you use the id as identifier:

let originalDocs = [
  {
    id: "sara",
    date: new Date("01/01/2022")
  },
  {
    id: "vic",
    date: new Date("10/26/2020")
  }
];

let newDocs = [
  {
    id: "vic",
    date: new Date("01/02/2022")
  },
  {
    id: "raul",
    date: new Date("05/05/2021")
  }
];
const map = new Map(originalDocs.map(el => [el.id, el]));
newDocs.forEach(el => map.set(el.id, el));
const sorted = [...map.values()].sort((a, b) => a.date - b.date);
console.log(sorted)

This overrides any originalDocs entry with a newDocs entry if they have the same id.

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