# Computing pairwise bitwise OR in a list of bit-strings

I am trying to solve this problem using Clojure:

You are given a list of $N$ people who are attending ACM-ICPC World Finals. Each of them are either well versed in a topic or they are not. Find out the maximum number of topics a 2-person team can know. And also find out how many teams can know that maximum number of topics.

Input Format:

The first line contains two integers $N$ and $M$ separated by a single space, where $N$ represents the number of people, and M represents the number of topics. $N$ lines follow. Each line contains a binary string of length $M$. In this string, 1 indicates that the ith person knows a particular topic, and 0 indicates that the ith person does not know the topic.

Output Format:

On the first line, print the maximum number of topics a 2-people team can know. On the second line, print the number of 2-person teams that can know the maximum number of topics.

Constraints:

$2 ≤ N ≤ 500$

$1 ≤ M ≤ 500$

Sample Input:

4 5

10101

11100

11010

00101


Sample Output:

5

2


However, my implementation is too slow to pass all the test cases. What are some suggestions to optimize this implementation? Basically, I have a list of bit-strings, and I am trying to find a faster way to bitwise-OR every pair of strings, and find the maximum number of set bits that could be obtained from ORing a pair.

(let [[N _] (clojure.string/split (read-line) #" ")
N (Integer/parseInt N)
known
(for [i (range N) j (range (inc i) N)
:let [s1 (strings i)
s2 (strings j)]]
(apply + (map #(if (or (= %1 \1)
(= %2 \1))
1
0)
s1
s2)))
maximum (apply max known)]
(println maximum)
(println (count (filterv #(= % maximum) known))))

• Please revise the title to state the challenge, not the review request. It would also help to embed the challenge objective. – Jamal Nov 9 '14 at 2:17
• "Finding the maximum bitwise union of pairs from a list of integers" is one possibility for a title that describes the challenge. Also, there's a bitwise tag that would be appropriate for this question. And consider tagging as lisp so that more people can find it. – Brythan Nov 9 '14 at 2:44

I was able to get it accepted by optimizing the inner loop by using BitSets instead of plain strings. I then just used the BitSet's or and bit counting method (cardinality) to do the work. Here's the resulting code:

(import [java.util BitSet])

(defn string->bitset [str]
(let [bitset (BitSet. (count str))]
(doall (map-indexed #(when (= \1 %2)
(.set bitset %1)) str))
bitset))

(defn bitset-or [s1 s2]
(let [tmp (.clone s1)]
(.or tmp s2)
tmp))

(let [[N _] (clojure.string/split (read-line) #" ")
N (Integer/parseInt N)
bitsets (mapv #(string->bitset %) (repeatedly N read-line))
known
(for [i (range N) j (range (inc i) N)
:let [s1 (bitsets i)
s2 (bitsets j)]]
(.cardinality (bitset-or s1 s2)))
maximum (apply max known)]
(println maximum)
(println (count (filterv #(= % maximum) known))))


One optimization would be to first build a map of the attendees with the number of topics for each one and find the maximum number of topics for a single person. You know that the maximum number for a pair will obvious be larger than the maximum for a single person. So when you loop over the pairs to compute the numbers of topics covered by a pair, you can skip the topics intersection computation if the sum of the topics for both attendees is less than the maximum.

I don't know that optimization alone will be sufficient.