Question

I saw this program on Codechef.
There are N packets each containing some candies. (Eg: 1st contains 10, 2nd contains 4 and so on)
We have to select exactly M packets from it ( M<=N) such that total candies are divisible by K.
If there are more than one solution then output the one having lowest number of candies.

I thought its similar to Subset Sum problem but that is NP hard. So it will take exponential time.

I don’t want the complete solution of this program. An algorithm would be appreciated. Thinking on it from 2 days but unable to get the correct logic.

1 ≤ M ≤ N ≤ 50000, 1 ≤ K ≤ 20
Number of Candies in each packet [1,10^9]

Answer 1

Let packets contain the original packets.

Partition k into sums of p = 1, 2, ..., m numbers >= 1 and < k (there are O(2^k) such partitions). For each partition, iterate over packets and add those numbers whose remainder modulo k is one of the partition’s elements, then remove that element from the partition. Keep the minimum sum as well, and update a global minimum. Note that if m > p, you must also have m - p zeroes.

You might be thinking this is O(2^k * n) and it’s too slow, but you don’t actually have to iterate the packets array for each partition if you keep num[i] = how many numbers have packets[i] % k == i, in which case it becomes O(2^k + n). To handle the minimum sum requirement too, you can keep num[i] = the list of the numbers that have packets[i] % k == i, which will allow you to always pick the smallest numbers for a valid partition.