[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 0Please use the above solution for reference.
I recommend you to write your own source code.
