[Codility]Lesson4/Counting Elements/permCheck

Mijeong Ryu·2023년 4월 23일
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Codility

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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].

코드

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.*;
class Solution {
    public int solution(int[] A) {
        Arrays.sort(A);
        int ans = 1;
        for(int i=0; i<A.length; i++){
            if(i+1 != A[i]){
                return 0;
            }
        }
        return ans;
    }
}

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