Codility PermCheck

int solution(int A[], int N) {
    
    int i = 0;
    int * array = calloc(N + 1, sizeof(int));
    int sum = N * (N + 1) / 2;
    int sum2 = 0;
    
    for (; i < N; i++) {
        
        if (A[i] <= N && A[i] > 0 && array[A[i]] == 0) {
         
            array[A[i]] = 1; 
            sum2 += A[i];
        }
    }
    
    free(array);
    return sum == sum2 ? 1 : 0;
}