Reputation: 764
Hello and sorry for lame title of this post, just couldn't find a better one.
So, I'm solving Codility exercise called NumberOfDiscIntersections. The solution requires some basic sorting and some minor arithmetic operations. I have achieved 93% result, and only one test is failing. The descripton that they provide is following:
For example, for the input [1, 2147483647, 0] the solution returned a wrong answer (got -1 expected 2).
Problem can be seen here.
And here is my solution:
typedef long long int my_type; //can't use unsigned!
#define LIMIT 10000000
//method used by qsort()
int comp(const void* left, const void* right) {
my_type arg1 = *(const my_type*)left;
my_type arg2 = *(const my_type*)right;
if(arg1 < arg2) return -1;
if(arg2 < arg1) return 1;
return 0;
}
int solution(int A[], int N) {
// write your code in C99 (gcc 6.2.0)
//allocate two arrays to hold beginning and ending points of each circle
my_type *lower = calloc(N, sizeof(my_type));
my_type *upper = calloc(N, sizeof(my_type));
int i;
my_type count = 0;
//initialize arrays
for(i = 0; i < N; i++) {
lower[i] = i - A[i];
upper[i] = i + A[i];
}
qsort(lower, N, sizeof(my_type), comp);
qsort(upper, N, sizeof(my_type), comp);
int open = 0;
int upper_index = 0;
for(i = 0; i < N; i++) {
while(lower[i] > upper[upper_index]) {
//printf("closing %d\n", upper[upper_index]);
upper_index++;
open--;
}
open++;
count += (open-1);
//printf("opening %d\n", lower[i]);
}
free(lower);
free(upper);
return ((int)count <= LIMIT) ? (int)count : -1;
}
Upvotes: 0
Views: 68
Reputation: 34583
The right hand side of these
lower[i] = i - A[i];
upper[i] = i + A[i];
performs int
addition. You must cast one of the operands:
lower[i] = (my_type)i - A[i];
upper[i] = (my_type)i + A[i];
to prevent integer overflow.
Upvotes: 2