Codility - Lesson 4 Counting elements - 2. PermCheck

Source Link:
https://codility.com/programmers/lessons/4-counting_elements/perm_check/

Question:

A non-empty zero-indexed 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 a zero-indexed 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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:

Notes:
1. To check "permutation", the main idea is as follows:
// 1. use set to remember which elements have appeared
// 2. use "for loop" to check if all the elements from "1 to A.length" appeared
// If all the elements have appeared, then "yes".
// Otherwise, "no".   
2.
Set<Integer> set = new HashSet<Integer>();
for(int i=0; i < A.length; i++){
set.add(A[i]);
}
3. check if "all" the elements from "1 to A.length" appeared
for(int i=1; i<= A.length; i++){
if( set.contains(i) == false )
return 0; // not a permutation (A[i] is missing)
}