Get indexes of substrings contained in a string in Kotlin-way

Question

I want to implement a function that will return the indexes of the substrings in the specified string. For now i did it in Java-style:

public fun String?.indexesOf(substr: String, ignoreCase: Boolean = true): List<Int> {
    var list = mutableListOf<Int>()
    if (substr.isNullOrBlank()) return list
    var count = 0;
    this?.split(substr, ignoreCase = ignoreCase)?.forEach {
        count += it.length
        list.add(count)
        count += substr.length
    }
    list.remove(list.get(list.size-1))
    return list
}

But I don't think this is a kotlin-way solution. Its most looks like typical java program but written in kotlin. How can this be implemented more elegantly using kotlin?

at least you can use list.dropLast(1) instead of list.remove(list.get(list.size-1)) — Andrei Tanana
– Andrei Tanana, Commented Jun 4, 2020 at 8:22

leetwinski · Accepted Answer · 2020-06-04 09:19:46Z

15

what would i do is the following:

fun ignoreCaseOpt(ignoreCase: Boolean) = 
    if (ignoreCase) setOf(RegexOption.IGNORE_CASE) else emptySet()

fun String?.indexesOf(pat: String, ignoreCase: Boolean = true): List<Int> =
    pat.toRegex(ignoreCaseOpt(ignoreCase))
        .findAll(this?: "")
        .map { it.range.first }
        .toList()

// check:
println("xabcaBd".indexesOf("ab", true))
println("xabcaBd".indexesOf("ab", false))
println("xabcaBd".indexesOf("abx", true))

val s: String? = null
println(s.indexesOf("aaa"))

// output:
[1, 4]
[1]
[]
[]

answered Jun 4, 2020 at 9:19

leetwinski

17.9k2 gold badges21 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Michael · Accepted Answer · 2020-06-04 09:16:07Z

9

You could condense it down to something like this:

public fun String?.indexesOf(substr: String, ignoreCase: Boolean = true): List<Int> {
    return this?.let { 
        val regex = if (ignoreCase) Regex(substr, RegexOption.IGNORE_CASE) else Regex(substr)
        regex.findAll(this).map { it.range.start }.toList()
    } ?: emptyList()
}

Whether that's more efficient is a different matter. You'd have to test that.

If you wanted "aaa".indexesOf("aa") to return [0, 1] rather than just [0], you should be able to do that by modifying the regex to use positive lookahead, i.e.:

val regex = if (ignoreCase) Regex("(?=$substr)", RegexOption.IGNORE_CASE) else Regex("(?=$substr)")

edited Jun 4, 2020 at 9:16

answered Jun 4, 2020 at 8:31

Michael

58.7k11 gold badges86 silver badges128 bronze badges

3 Comments

gidds Over a year ago

@IR42 That means exactly the same. (They both return the EmptyList singleton object.)

shantanu Over a year ago

If the substring contains Parentheses then it returns 0 occurrences.

gswierczynski Over a year ago

Make sure to escape some of the characters. In my case I had to escape $ sign by .replace("$","\\$") for this to work

Nikolai Shevchenko · Accepted Answer · 2020-06-04 08:48:23Z

4

Correct way is to use String.indexOf(), since splitting would ignore some substring occurrences.

For example with input "aaaa" and substr "aaa" ("aaaa".indexesOf("aaa")) result should be [0, 1] but your solution (using split) will result to [0]

public fun String?.indexesOf(substr: String, ignoreCase: Boolean = true): List<Int> {
    val list = mutableListOf<Int>()
    if (this == null || substr.isBlank()) return list

    var i = -1
    while(true) {
        i = indexOf(substr, i + 1, ignoreCase)
        when (i) {
            -1 -> return list
            else -> list.add(i)
        }
    }
}

edited Jun 4, 2020 at 8:48

answered Jun 4, 2020 at 8:41

Nikolai Shevchenko

7,6278 gold badges36 silver badges47 bronze badges

Comments

mudlej · Accepted Answer · 2022-08-24 11:42:53Z

This should be a comment leetwinski's anseer , but SO won't let me write comments.

It's a great solution but be aware that if your query string contains any special characters that have special meaning in regex, that may give you incorrect results or even a PatternSyntaxException and crash your app.

So, if you want to look for a literal match, you have to use escape

So the code will

fun ignoreCaseOpt(ignoreCase: Boolean) =
        if (ignoreCase) setOf(RegexOption.IGNORE_CASE) else emptySet()

fun String?.indexesOf(query: String, ignoreCase: Boolean = true): List<Int> =
    Regex.escape(query)       // to disable any special meaning of query's characters
        .toRegex(ignoreCaseOpt(ignoreCase))
        .findAll(this?: "")
        .map { it.range.first }
        .toList()

Jakub Zalas · Accepted Answer · 2020-06-04 13:24:13Z

3

Here's a tail-recursive example that doesn't hold any mutable state:

fun String?.indexesOf(substr: String, ignoreCase: Boolean = true): List<Int> {
    tailrec fun String.collectIndexesOf(offset: Int = 0, indexes: List<Int> = emptyList()): List<Int> =
        when (val index = indexOf(substr, offset, ignoreCase)) {
            -1 -> indexes
            else -> collectIndexesOf(index + substr.length, indexes + index)
        }

    return when (this) {
        null -> emptyList()
        else -> collectIndexesOf()
    }
}

"abcABCbcaabcabcaaabc".indexesOf("ddd")
// []
"abcABCbcaabcabcaaabc".indexesOf("abc", ignoreCase = false)
// [0, 9, 12, 17]
"abcABCbcaabcabcaaabc".indexesOf("abc", ignoreCase = true)
// [0, 3, 9, 12, 17]
null.indexesOf("abc", ignoreCase = true)
// []

It will find the first index of the substring, and recursively continue shortening it to find the next occurrence.

edited Jun 4, 2020 at 13:24

answered Jun 4, 2020 at 9:04

Jakub Zalas

36.3k10 gold badges98 silver badges127 bronze badges

3 Comments

Nikolai Shevchenko Over a year ago

Each call substring(index + substr.length) will produce new string. That's not efficient, especially on large strings

Jakub Zalas Over a year ago

I'd love to see your benchmarks ;) this question isn't about micro-optimisations people.

Jakub Zalas Over a year ago

@IR42 that's a nice improvement. I will update the answer.

ATutorMe · Accepted Answer · 2021-12-16 08:13:07Z

3

I really like @leetwinski's and @Michael's answers.

There's so many possibilities with Kotlin, it's amazing :)

Another possible solution based on the above:

fun String.indexesOf(substr: String, ignoreCase: Boolean = true) : List<Int> =
    (if (ignoreCase) Regex(substr, RegexOption.IGNORE_CASE) else Regex(substr))
        .findAll(this).map { it.range.first }.toList()

@JvmName("indexesOfNullable")
fun String?.indexesOf(substr: String, ignoreCase: Boolean = true) = this?.indexesOf(substr, ignoreCase) ?: emptyList()

answered Dec 16, 2021 at 8:13

ATutorMe

9448 silver badges15 bronze badges

1 Comment

ATutorMe Over a year ago

PS: the @JvmName("indexesOfNullable") annotation is because the JVM method signatures clash. To see the full message remove the annotation in IntelliJ IDEA.

shantanu · Accepted Answer · 2021-02-07 08:30:35Z

1

Try this using the indexOf function

fun String?.indexesOf(substr: String, ignoreCase: Boolean = false): List<Int> {
    return this?.let {
        val indexes = mutableListOf<Int>()
        var startIndex = 0
        while(startIndex in 0 until length){
            val index = this.indexOf(substr, startIndex, ignoreCase)
            startIndex = if(index != -1){
                indexes.add(index)
                index + substr.length
            } else{
                index
            }
        }
        return indexes
    } ?: emptyList()
}

answered Feb 7, 2021 at 8:30

shantanu

2,4182 gold badges32 silver badges58 bronze badges

Collectives™ on Stack Overflow

Get indexes of substrings contained in a string in Kotlin-way

7 Answers 7

Comments

3 Comments

Comments

Comments

3 Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

Comments

3 Comments

Comments

Comments

3 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related