solution.kt

package com.igorwojda.string.getduplicatedarguments

// Time complexity: O(n)
// Determine the frequency of each argument and then filter arguments with frequency > 1
private object Solution1 {
    fun getDuplicatedArguments(vararg strings: String): List<String>? = strings
        .groupingBy { it }
        .eachCount()
        .filter { it.value != 1 }
        .map { it.key }
}

// Time complexity: O(n)
// Optimal solution using double pointer.
private object Solution2 {
    fun getDuplicatedArguments(vararg strings: String): List<String>? {
        var pointer1 = 0
        var pointer2 = 1
        val result = mutableSetOf<String>()
        val sortedList = strings.sorted()

        for (it in 0 until sortedList.lastIndex) {
            if (sortedList[pointer1] == sortedList[pointer2]) {
                result.add(sortedList[pointer1])
            }

            pointer1++
            pointer2++
        }

        return result.toList()
    }
}

// Time complexity: O(n^2)
// Loop through all arguments and compare given argument with all other arguments.
private object Solution3 {
    fun getDuplicatedArguments(vararg strings: String): List<String>? {
        val result = mutableSetOf<String>()

        strings.forEachIndexed { index, element1 ->
            (index + 1..strings.lastIndex).forEach { index2 ->
                if (element1 == strings[index2]) {
                    result.add(element1)
                }
            }
        }

        return result.toList()
    }
}