Permutation Array Check in Python and C++ Codility Solutions

This video describes the permutation permcheck lesson 4 of codility programming interview exercises, in C++ and in Python. A permutation array is described and 2 algorithms are detailed showing the difference with the efficient solution. Hope you will enjoy learning this example. python algorithms interview.

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[wonderjaxe]

Great explanation as usual.
This one was surprisingly easy given the nightmare FrogRiverOne problem that comes before it.
My solution uses set theory to do the comparison, but pretty much identical approach:

def solution(A):
n = len(A)
A = set(A)
B = set(x for x in range(1, n+1))
if len(A&B) == n:
return 1
return 0

[juandavidtorres]

My solution, using what I learned from the book "Python One-Liners"

def solution(A):
return 1 if sorted(A) == [*range(1, len(A)+1)] else 0

[TorstigHjalmr]

It doesn't mention whether duplicates are sent in. Convert to set, then sort, then compare the first and last items could result in a quicker algorithm.

[TonyBoy]

Hey, great videos! Question though, does the sorting not increase the complexity of O(n)?

[confluenceking]

Would it be right to order the array first and then see that for each element starting from index 0 if the next element is the current element + 1 then return true, else return false?

[gamesniper]

How is your solution O(n)? Youre using a sort method, none of which are linear.