# [Codility] Lesson-04.4: PermCheck

The topic of this post is checking the permutation. The problem that we want to solve follows.

# Task Description

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:

def solution(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].

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# Key Point

- The sorting algorithm is not needed.
- The xor operator makes it easy to check the existence of each permutation.

# Solution (using Python)

**def** solution(A):

A.**extend**(**list**(**range**(1, **len**(A)+1)))

xor = 0

**for** val **in** A: xor ^= val

**if**(xor == 0): **return** 1

**else**: **return** 0

Please use the above solution for reference.

I recommend you to write your own source code.