[Codility][Kotlin]PermCheck

문제

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:

fun solution(A: IntArray): Int

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].

 

풀이

fun solution(A: IntArray): Int {
    val a = A.size
    val seen = BooleanArray(A.size+1)
    A.forEach{
        if(it > a || seen[it]){
            return 0
        }
        if(!seen[it]){
            seen[it] = true
        }
    }
    return 1
}

무난하게 풀이 성공~