Problema:

A non-empty zero-indexed 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:

int solution(int A[], int N);

that, given a zero-indexed 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.

Assume that:

  • 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].

Complexity:

  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Solución:


def solution(A):
    if set(A) == set(xrange(1, len(A)+1)):
        return 1
    return 0

En este problema nos dice que dada una lista, retornemos 1 (uno) si es una permutacion o 0 (cero) si no lo es (una permutacion es un conjunto de elementos (en este caso números de 1 a N) donde ningún valor se repite), por lo tanto en mi solución hago uso de set, set toma un iterable y retorna un objeto set que tiene varios métodos y operaciones (==, <, <=, >, =>, union, intesection, symmetric_difference, etc…):

  • Convierto A en un set
  • Creo un set de 1 a la longitud de A
  • Comparo ambos sets
  • Retorno 1 si ambos sets son iguales
  • Si no son iguales retorno 0

Aquí esta el repo con todas mis soluciones de codility: https://github.com/OmarIbannez/codility-lessons

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