codility_tasks/perm_check.py at master · oluwasayo01/codility_tasks · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
"""
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:

class Solution { public int solution(int[] 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].
"""


def solution(A):
    actual_sum = sum(range(len(A) + 1))
    list_sum = sum(A)
    actual_length = len(set(A))
    if (list_sum == actual_sum) and (actual_length == len(A)):
        return 1
    return 0

print(solution([4, 1, 3, 2]))
print(solution([4, 1, 3]))