[Codility] Lesson4 - PermCheck

Task discription

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].
  
  

Source code

#include <iostream>
#include <algorithm>
#include <vector>

int solution(vector<int> &A) {
    sort(A.begin(), A.end());

    int num = 1;
    for(int i=0;i<A.size();i++) {
        if(num != A[i]) {
            return 0;
        }
        num++;
    }
    return 1;
}