Lesson 4 - PermCheck

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].

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내답안

Set을 활용해서 중복 값을 포함하는지 검사했다. 흠 근데 찝찝하다 굳이 이렇게 풀 필요는 없을 것 같았다.

import java.util.HashSet;
import java.util.Set;
class Solution {
    public int solution(int[] A) {
        boolean isPermutation = false;
        boolean isDistinctValueExisted = false;

        Set permutation = new HashSet();
        int maxLength = 0;
        for(int i: A) {
            if (maxLength < i) {
                maxLength = i;
            }
            
            if (permutation.contains(i)) {
                isDistinctValueExisted = true;
                break;
            } else {         
                permutation.add(i);
            }           
        }

        if (permutation.size() == maxLength && !isDistinctValueExisted) {
            isPermutation = true;
        }                
        
        return isPermutation ? 1 : 0;
    }
}

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좋은 답안

정열을 시키고 숫자를 1증가 시켜 비교하였다. 순열이 나오면 정렬하는걸 잊지 말아야겠다.

public int solution(int[] A) {
    int len = A.length;
    Arrays.sort(A);
    for(int i = 0; i < len; i++){
        if(A[i] != i+1){
            return 0;
        }
    }
    return 1;
}