Question

So I was developing an algorithm to count the number of repetitions of each character in a given word. I am using a HashMap and I add each unique character to the HashMap as the key and the value is the number of repetitions. I would like to know what the run time of my solution is and if there is a more efficient way to solve the problem.

Here is the code :

public static void getCount(String name){
        public HashMap<String, Integer> names = new HashMap<String, Integer>() ;
        for(int i =0; i<name.length(); i++){

            if(names.containsKey(name.substring(i, i+1))){
                names.put(name.substring(i, i+1), names.get(name.substring(i, i+1)) +1);
            }
            else{
                names.put(name.substring(i, i+1), 1);
            }
        }

        Set<String> a = names.keySet();
        Iterator i = a.iterator();

        while(i.hasNext()){
            String t = (String) i.next();
            System.out.println(t + " Ocurred " + names.get(t) + " times");
        }
    }

Answer 1

The algorithm has a time complexity of O(n), but I’d change some parts of your implementation, namely:

• Using a single get() instead of containsKey() + get();
• Using charAt() instead of substring() which will create a new String object;
• Using a Map<Character, Integer> instead of Map<String, Integer> since you only care about a single character, not the entire String:

In other words:

public static void getCount(String name) {
    Map<Character, Integer> names = new HashMap<Character, Integer>();
    for(int i = 0; i < name.length(); i++) {
        char c = name.charAt(i);
        Integer count = names.get(c);
        if (count == null) {
            count = 0;
        }
        names.put(c, count + 1);
    }
    Set<Character> a = names.keySet();
    for (Character t : a) {
        System.out.println(t + " Ocurred " + names.get(t) + " times");
    }
}