Lonely Integer

Given an array of integers, where all elements but one occur twice, find the unique element.

Example a=(1,2,3,4,2,1)

The unique element is 4.

Function Description

Complete the lonelyinteger function in the editor below.

lonelyinteger has the following parameter(s):

int afny: an array of integers

Returns

• int the element that occurs only once

Input Format

The first line contains a single integer, n, the number of integers in the array.

The second line contains n space-separated integers that describe the values in a. Constraints

• 1 ⇐ n < 100

• It is guaranteed that m is an odd number and that there is one unique element.

• 0 ⇐ a[i] ⇐ 100, where O ⇐ i < n. 

fun lonelyinteger(a: Array<Int>): Int {
    val maxSize = 100
    val repetitionsArray = Array<Int>(maxSize){0}
    for (index in a.indices) {
        val newIndex = a[index] % maxSize
        repetitionsArray[newIndex] += 1
    }
    for (index in repetitionsArray.indices) {
        if(repetitionsArray[index] == 1){
            return index
        }
    }
    return 0
}

Solution #2

fun lonelyinteger(a: Array<Int>): Int {
    val numberMap = HashMap<Int, Int>()
    for (number in a){
        val currentQuantity = numberMap[number] ?: 0
        numberMap[number] = currentQuantity + 1
    }
    for (number in numberMap) {
        if(number.value == 1){
            return number.key
        }
    }
    return -1