Kotlin - index of duplicate elements in an array

Question

Is there an easy way to get the index of matching elements in an array without writing a for loop and then collecting the index?

Eg: val arr = arrayOf<Int>(2,3,4,2,5,2,6,3,2)

Output: For element 2, output should be (0,3,5,8) and For element 3, output should be (1,7)

Kotlin provides indexOf(element) which returns first index and then lastIndexOf(element) which will give me last index. Am looking for an easy way to get indexes of all matching elements.

My solution: Standard solution of iterating over the array and collecting index.

var indexArrOf2 = mutableListOf<Int>()
var indexArrOf3 = mutableListOf<Int>()
for(i in arr.indices) {
  if (arr[i] == 2) {
    indexArrOf2.add(i)
  }
  if (arr[i] == 3) {
    indexArrOf3.add(i)
  }
}

This may not be efficient at all, but you could do something like this: arr.mapIndexed { index, value -> if (value == 2) index else null }.filterNotNull() — Michael
– Michael, Commented Apr 5, 2020 at 8:03

Giorgio Antonioli · Accepted Answer · 2020-04-05 09:19:03Z

9

Yes, you can use mapIndexedNotNull:

arr.mapIndexedNotNull { index, elem -> if (elem == 2) index else null }

answered Apr 5, 2020 at 9:19

Giorgio Antonioli

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Comments

Sourav 'Abhi' Mitra · Accepted Answer · 2020-04-06 05:48:27Z

1

A little convoluted but returns a map Map<Element, List<Indices>> to use as desired.

arr.mapIndexed { index, i -> i to index  }   //List<Pair<Int, Int>> - [(2, 0), (3, 1), (4, 2), (2, 3), (5, 4), (2, 5), (6, 6), (3, 7), (2, 8)]
.groupBy { it.first }   //Map<Int, List<Pair<Int, Int>>> - {2=[(2, 0), (2, 3), (2, 5), (2, 8)], 3=[(3, 1), (3, 7)], 4=[(4, 2)], 5=[(5, 4)], 6=[(6, 6)]} 
.mapValues { it.value.map { it.second } }   //Map<Int, List<Int>> - {2=[0, 3, 5, 8], 3=[1, 7], 4=[2], 5=[4], 6=[6]}

edited Apr 6, 2020 at 5:48

answered Apr 6, 2020 at 5:34

Sourav 'Abhi' Mitra

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Comments

gidds · Accepted Answer · 2020-04-05 08:14:09Z

0

I'd use the withIndex() extension function to add the index. You can then select the matching items, and get their indices:

val arr = arrayOf(2, 3, 4, 2, 5, 2, 6, 3, 2)

println(arr.withIndex().filter{ it.value == 2 }.map{ it.index })
// prints '[0, 3, 5, 8]'

answered Apr 5, 2020 at 8:14

gidds

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1 Comment

Animesh Sahu Over a year ago

I think filter and map will loop (total loop will be twice), this is not ideal its even less optimized than original OP's approach i'd suggest using sequences as this will loop through only one.

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