Codility - PermCheck

Codility - PermCheck 순열 체크하는 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:

function solution(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].

 

Solution

function solution(A) {
    const len = A.length;
    if(len < 1 || len > 100000) {
        return 0;
    }
    const aSum = len*(len+1) / 2; // 현재 A의 길이의 대한 총합
    const distinctA = new Set(A); // 중복제거(순열이라면 1 ~ N 까지 한 번만 있을것)
    const aRealSum = [...distinctA].reduce((acc, cur) => acc + cur); // 중복 제거후 각 요소에 대한 전체 합
    return aSum === aRealSum ? 1 : 0;
}