Codility solutions - lesson2, exercise: PermCheck

Check whether array A is a permutation.

Content of Codility exercise:

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:

class Solution { public int solution(int[] A); }

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.

Solution:

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public int solution(int[] A) { int N = A.length; //create tmp array to store information about each number occurrence int tmbB[] = new int[A.length]; int element = 0; for (int i = 0; i < A.length; i++){ element = A[i]; //if element is greater than array A length it is obvious that A isn't permutation if (element> N){ return 0; } tmbB[element-1]++; } //index where number in array is equal zero, is our missing element //and array A isn't permutation if (isContainZeroElement(tmbB)){ return 0; }else { return 1; } } private boolean isContainZeroElement(int [] B){ for (int element: B){ if (element == 0){ return true; } } return false; }