Question

Given a sorted list of something (a,a,b,c,c)

What would be the most efficient way to recognize that a exists in the list 2 times, b once and c 2 times?

Aside from obvious making a map of counts. Can we do better then this?

            if (map.containsKey(key)) {
                map.put(key, map.get(key) + 1);
            } else {
                map.put(key, 1);
            }

Ultimately the goal is to iterate of the list and know at any given point how many times a key was seen before. Putting things in a map, seems like a step we don’t really need.

Answer 1

Your method, at each iteration, makes

• one lookup for containsKey
• one lookup for get
• one unboxing from Integer to int
• one boxing from int to Integer
• one put

You could simply compare the current element to the previous one, increment a count if it’s equal, and put the count if not (and reset the counter to 1).

But even if you keep your algorithm, using get and compare the result to null would at least avoid an unnecessary lookup.