Perm-Check with python

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:

def solution(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.


Solution

def solution(A):

    counter = [0]*len(A)

    limit = len(A)

    for element in A:

        if not 1 <= element <= limit:

            return 0

        else:

            if counter[element-1] != 0:

                return 0

            else:

                counter[element-1] += 1

 

    return 1