Imagine we have array like this:

const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];

Now I would like to sort this array alphabetically:

words.sort((a, b) => a.localeCompare(b));

Imagine we have array of super words. This words have high priority and should be on the top of 'words list'.

const superWords = ["zac1", "Test"]

So the desired result of words array will be:

const desiredResult = ["Test", "zac1","Alf0", "Bar", "luigi"]

How can I do it?

CodePudding user response：

You could convert superWords into a Set for O(1) lookup time using has rather than O(n) using includes when superWords is an Array (where n is the number of words in superWords):

const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];
const superWords = ['zac1', 'Test'];
const superWordsSet = new Set(superWords); // This is O(n) but you only need to do this once.
words.sort((a, b) => {
    if (superWordsSet.has(a) && !superWordsSet.has(b)) {
        return -1;
    } else if (!superWordsSet.has(a) && superWordsSet.has(b)) {
        return 1;
    }
    return a.localeCompare(b);
});
console.log(words);

Performance

The Set has method checks if a value is in a Set object, using an approach that is, on average, quicker than testing most of the elements that have previously been added to the Set object. In particular, it is, on average, faster than the Array.prototype.includes method when an Array object has a length equal to a Set object's size.

CodePudding user response：

you can do this

const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];


words.sort((a, b) => {
 const superWords = ["zac1", "Test"]
 if(superWords.includes(a) && !superWords.includes(b)){
  return -1
 }else if(superWords.includes(b) && !superWords.includes(a)){
  return 1
 }
 return a.localeCompare(b)

})

console.log(words)

CodePudding user response：

First step find doubles and the your can sort both arrays and afterwards you can concat the arrays.

let words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];
const superWords = ["zac1", "Test"]

words = words.filter(w => ! superWords.includes(w));

words.sort((a, b) => a.localeCompare(b));
superWords.sort((a, b) => a.localeCompare(b));



console.log(superWords.concat(words));

CodePudding user response：

This could help. You can click on 'Run code snippet' to see output:

const superWords = ['zac1', 'Test'];

const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];

const desiredResult = superWords.sort((a, b) => a.localeCompare(b));

desiredResult.push(
  ...words
  .filter((word) => !superWords.includes(word))
  .sort((a, b) => a.localeCompare(b))
);

console.log(desiredResult);