[codillity] kotlin - lesson4. PermCheck

A non-empty array A consisting of N integers is given.

A permutation is a sequence containing each element from 1 to N once, and only once.

For example, array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

is a permutation, but array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

is not a permutation, because value 2 is missing.

The goal is to check whether array A is a permutation.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A, returns 1 if array A is a permutation and 0 if it is not.

For example, given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

the function should return 1.

Given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

the function should return 0.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [1..1,000,000,000].

private fun solution(A: IntArray): Int {

    val list = mutableSetOf<Int>()
    A.forEach {
        //1차 걸러내기 사이즈보다 크면 연결 될 수 없음
        if(it > A.size) return 0

        //2차 걸러내기 중복값이 있으면 연결 될 수 없음
        if(list.contains(it)) return 0

        list += it
    }
    //끝나면 리턴
    return 1
}