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assuming the numbers are unsorted.
- netta October 29, 2016first we'll use the 10k space as a sort of hash table (HT).
the hash function will be a simple mod 10000.
for each number n in the list increment the appropriate cell in HT: HT[n%10000]++;
after going over the whole list, save the hash table cells that have less than 10000 in the solution array (i'm assuming we get an additional array or list for the solution).
then for each of those numbers N zero the 10K array, make another pass on the 10B list and save all the numbers where modulo 10000 equals N.
we need O(N) passes on the list where N is the number of missing numbers.