int main(int argc, char *argv[]) {
char *prog = argv[0];
int opt, rc=-1;
hex_t h;
utarray_new(cfg.hexv, &hex_icd);
while ( (opt = getopt(argc, argv, "v+i:o:h")) != -1) {
switch (opt) {
case 'v': cfg.verbose++; break;
case 'i': cfg.infile = strdup(optarg); break;
case 'o': cfg.outfile = strdup(optarg); break;
case 'h': default: usage(prog); break;
}
}
if (!cfg.outfile || !cfg.infile) usage(prog);
/* read input file */
tpl_node *tn = tpl_map("A(sii)", &h.id, &h.x, &h.y);
if (tpl_load(tn, TPL_FILE, cfg.infile)) goto done;
while(tpl_unpack(tn,1) > 0) utarray_push_back(cfg.hexv, &h);
tpl_free(tn);
find_bounds();
draw_image();
rc = 0;
done:
utarray_free(cfg.hexv);
return rc;
}
void findFrequency()
{
/*
@Statement - Given a sorted array, find the number of occurances of a number in less than O(n)
*/
int array[7] = {1,1,2,3,3,4,5};
int val = 4;
find_bounds(array,7,val);// use binary search --O(logn)
}