[algorithm] permutation check

mgm-dev·2021년 4월 21일
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codility algorithms

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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].


my solution

def solution(A):
    # write your code in Python 3.6
    A =  sorted(A)
    length = len(A) 
    if length != A[length-1]:
        return 0
    for i, v in enumerate(A):
        if i + 1 != v:
            return 0
    return 1

Detected time complexity: O(N) or O(N * log(N))

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