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 = 4

A = 1

A = 3

A = 2

is a permutation, but array A such that:

A = 4

A = 1

A = 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 = 4

A = 1

A = 3

A = 2

the function should return 1.

Given array A such that:

A = 4

A = 1

A = 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].

# 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.

## More from YeongHyeon Park

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