[Codility]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].

 

- 배열의 수가 순열이면 1 아니면 0을 리턴

1. 첫번째 시도

트리셋에 넣어서 순차 정렬 한 후 찾아내려고 함...

테스트 케이스 빼고 다 실패....ㅠㅠ

// 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");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        TreeSet<Integer> treeSet = new TreeSet<Integer>();
        int returnValue = 0;

        for(int i:A){
            treeSet.add(i);
        }
        int num = treeSet.first();
        while(treeSet.contains(num)){
            if(num == treeSet.last()){
                returnValue = 1;
                break;
             }
            num++;
        }
        
        return returnValue;
    }
}

 

 

2. 두번째 시도

// 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");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
       // TreeSet<Integer> treeSet = new TreeSet<Integer>();

        int returnValue = 0;
        int[] sortA = new int[A.length];
        Arrays.sort(A);
        int minValue = A[0];
        for(int i=0 ; i<A.length ; i++){
            if(A[i]  == minValue){
                returnValue = 1;
            }else{
                returnValue = 0;
                break;
            }
            minValue ++;
        }
        return returnValue;
    }
}

 

 

단일 값에 대한 처리 및 시작값이 1이 아닌값에 대한 처리가 부족!

 

 

3. 최종버전

// 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");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
       // TreeSet<Integer> treeSet = new TreeSet<Integer>();

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