How to sort duplicate element based on index position in javascript

Question

Input :

arr[] = [3, 4, 3, 5, 2, 3, 4, 3, 1, 5]

After sorting array by, array as shown below

[1(8), 2(4), 3(0), 3(2), 3(5), 3(7), 4(1), 4(6), 5(3), 5(9)]

The expected output is invalid. Please create a [mcve] – adiga Jun 23 '19 at 11:46 — adiga, Jun 23 '19 at 11:46

Ziv Ben-Or · Answer 1 · 2019-06-23T11:09:38.233

0

You can map the values with the index. Then sort, and finally map to the desire string:

var arr = [3, 4, 3, 5, 2, 3, 4, 3, 1, 5];

console.log(arr
   .map((x, i) => [x, i])
   .sort((a, b) =>  a[0] - b[0])
   .map(x => `${x[0]}(${x[1]})`));

edited Jun 23 '19 at 11:09

answered Jun 23 '19 at 10:57

Ziv Ben-Or

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Just `sort` isn't enough, try adding `10` to the array. – georg Jun 23 '19 at 10:58
@georg, good point! I edited the answer. – Ziv Ben-Or Jun 23 '19 at 11:01

score 0 · Answer 2 · answered Jun 23 '19 at 11:02

0

Try this:

const a = [3, 4, 3, 5, 2, 3, 4, 3, 1, 5]
const result =a.map((el, i) => ({ index: i, value: el }))
.sort( (a, b) => a.value - b.value)
.map(({index,value})=>` ${value}(${index})`);

console.log(result)

answered Jun 23 '19 at 11:02

Ghoul Ahmed

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How to sort duplicate element based on index position in javascript

2 Answers2