Get array of array numbers

Question

I have an array like this one:

let array = [14, 42, 1, 3]

And I would like to get the arrays number mapped to this:

[1, 0, 3, 2]

Here is the reason:

1: because 14 is the second biggest number
0: because 42 is the biggest number
3: ...

What I have tried so far:

let sort = (array) => {
  let result = []
  let x = array.slice(0).sort((a, b) => b - a)
  for (let elem of x) {
    result.push(array.indexOf(elem))
  }
  console.log(result)
}

// Working
sort([14, 42, 1, 3]) // [1, 0, 3, 2]

// Not working, includes the index "0" two times
sort([14, 42, 14, 3]) // [1, 0, 0, 3]
// Expected: [1, 0, 2, 3]

Something like [ 1, 0, 2, 3]

Jonas
– Jonas

2019-07-09 18:14:37 +00:00
Commented Jul 9, 2019 at 18:14 — Jonas
– Jonas, Commented Jul 9, 2019 at 18:14
can you explain the logic a little bit?

Manos Kounelakis
– Manos Kounelakis

2019-07-09 20:08:21 +00:00
Commented Jul 9, 2019 at 20:08 — Manos Kounelakis
– Manos Kounelakis, Commented Jul 9, 2019 at 20:08

Nina Scholz · Accepted Answer · 2019-07-09 18:20:39Z

9

You could take the indices and sort them by taking the value from the given array.

const sort = array => [...array.keys()].sort((a, b) => array[b] - array[a]);

console.log(sort([14, 42, 1, 3]));
console.log(sort([14, 42, 14, 3]));

answered Jul 9, 2019 at 18:20

Nina Scholz

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1 Comment

Hasan Patel Over a year ago

this is a cool solution, thanks this will help me in the future with similar cases.

Alex Szeliga · Accepted Answer · 2019-07-09 18:24:29Z

2

It's because indexOf stops when it finds it's first result.

You could try to change the value to null once it's located the first time, or compare value to values already in the result and ignore those values.

let sort = (array) => {
  let result = []
  let x = array.slice(0).sort((a, b) => b - a)
  for (let elem of x) {
    result.push(array.indexOf(elem))
    array[array.indexOf(elem)] = null;
  }
  console.log(result)
}

answered Jul 9, 2019 at 18:24

Alex Szeliga

172 bronze badges

Comments

David Kirk · Accepted Answer · 2019-07-09 18:22:51Z

0

let sort = (arr) => {
    let arr2 = arr.slice().sort((a, b) => b - a);
    return arr.map((val) => {
        return arr2.indexOf(val);
    })
}
console.log(sort([14, 42, 1, 3]));

answered Jul 9, 2019 at 18:22

David Kirk

3423 silver badges5 bronze badges

Comments

Code Maniac · Accepted Answer · 2019-07-09 18:27:45Z

0

You can use a tracker object, map value and indexes as key/value pair and when looping though the array take the first index from respective key and shift it as well

let sort = (array) => {
  let result = []
  let x = array.slice(0).sort((a, b) => b - a)
  let tracker = x.reduce((op,inp,i)=>{
    op[inp] = op[inp] || []
    op[inp].push(i)
    return op
  },{})
  
  for (let elem of array) {
    let val = tracker[elem][0]
    tracker[elem].shift()
    result.push(val)
  }
  console.log(result)
}

sort([14, 42, 1, 3]) // working

sort([14, 42, 14, 3]) // includes the index "0" two times

answered Jul 9, 2019 at 18:27

Code Maniac

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Comments

Brian Swisher · Accepted Answer · 2019-07-09 18:49:25Z

0

(() => {

  function getBiggestOrder (nums) {
    const lookup = {}
    const result = nums.slice(0).sort((a, b) => b - a).map((num, i) => {
      lookup[num] = i
      return num
    })
    
    return nums.map(n => lookup[n])
  }

  const op = getBiggestOrder([14, 42, 1, 3])

  console.log(op)

  return op

})()

answered Jul 9, 2019 at 18:49

Brian Swisher

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Comments

Khizar Muhammad · Accepted Answer · 2019-07-13 05:19:19Z

0

You are basically numbering the numbers from biggest to smallest. sort them in a duplicate array from biggest to smallest. And replace the original array numbers with their index in the duplicate array.

Original = [14, 42, 1, 3]
Duplicate = [42, 14, 3, 1]
Duplicate indexes are [0, 1, 2, 3]

so find 42 in the first array and replace it with the index of the 42 in the duplicate array, etc.

answered Jul 13, 2019 at 5:19

Khizar Muhammad

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