[codility] 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:

public func solution(_ A : inout [Int]) -> 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].
  
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문제 해설

A배열이 순열인지 아닌지 체크 

순열이면 1 아니면 0 리턴

 

구현 해설

A 배열이 [3,1,2,4] 인경우에는 1 리턴

  [1,2,4] 인 경우에는 순열이 아니기 때문에 0 리턴

 

import Foundation
import Glibc

// you can write to stdout for debugging purposes, e.g.
// print("this is a debug message")

public func solution(_ A : inout [Int]) -> Int {
    // write your code in Swift 4.2.1 (Linux)

    A.sort()
    for (index, _) in A.enumerated() {
        if index+1 != A[index] {
            return 0
        }
    }
    return 1
}