Lesson4_1(PermCheck) - WhyAbout/CodilityAlgorithm GitHub Wiki

Task description
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(vector<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.

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    unsigned int l = A.size();
    vector<int> permArr(l, 0);
    int isPermArray = 1;
    for(unsigned int i = 0; i < l; ++i){
        if( (unsigned int)A[i] > l || permArr[A[i]-1] != 0 ) {
            isPermArray = 0;
            break;
        }
        permArr[A[i]-1] = 1;
    }
    return isPermArray;
}