[Codility] Lesson 4 - Counting Elements : PermCheck

Lesson 4 - Counting Elements : 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:

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].

📝 풀이

• permutation 일 경우 A의 길이와 제일 큰 숫자가 같음
• A 의 길이보다 작으면서 해당 숫자를 포함하고 있는지 여부 확인
• 1부터 N까지 포함여부를 확인해서 하나라도 포함 안 된 경우가 있는지 확인

💻 구현

static class Solution{
    public int solution(int[] A){
        int answer = 1;
        boolean[] contains = new boolean[A.length + 1];

        for (int i = 0; i < A.length; i++) {
            int number = A[i];

            if(number <= A.length && !contains[number]){
                contains[number] = true;
            }
        }

        for (int i = 1; i < contains.length; i++) {
            if(!contains[i]){
                return 0;
            }
        }

        return answer;
    }
}