Counting Elements - PermCheck

Question

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].

Test Result

public int solution(int[] A) {
        Set<Integer> nums = new HashSet<Integer>();
        for (int i = 1; i <= A.length; i++) {
            nums.add(i);
        }

        for (int a : A) {
            if (!nums.contains(a)) {
                return 0;
            } else {
                nums.remove(a);
            }
        }
        return nums.isEmpty() ? 1 : 0;
    }