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Codility 7번 PermCheck 본문

개발/알고리즘

Codility 7번 PermCheck

룬아님 2019. 8. 12. 21:37

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].

 

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) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();

        for(int i=0; i<A.length; i++){
            map.put(A[i], 1);
        }
        
        for(int i=0; i<A.length; i++){
            if(!map.containsKey(i+1)){
                return 0;
            }
        }
        
        return 1;
        // write your code in Java SE 8
    }
}

https://app.codility.com/demo/results/trainingWQF79W-39X/

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