Turn String aaaabbbbffffd into a4b4d3

Question

I\'m trying to get a head start on practicing interview questions and I came across this one:

Turn String aaaabbbbffffd into a4b4d3

You would basically want t

Answer 1

To add on to @Makoto's wonderful answer, in your situation, I would use a TreeMap instead of a HashMap. A TreeMap will allow you to print in alphabetical order. I have also added the print code to show you how it would look. It's fully runnable.

import java.util.Map;
import java.util.TreeMap;

public class MapPractice {

    public static void main(String[] args) {
        Map<Character, Integer> map = new TreeMap<>();

        String blah = "aaaabbbbffffd";

        for (int i = 0; i < blah.length(); i++) {
            char c = blah.charAt(i);
            if (!map.containsKey(c)) {
                map.put(c, 1);
            } else {
                map.put(c, (map.get(c) + 1));
            }
        } 

        for (Map.Entry<Character, Integer> entry: map.entrySet()) {
            System.out.print(entry.getKey() + "" + entry.getValue());
        }
    }
}

Output with TreeMap: a4b4d3

Output with HashMap: d3b4a4

Answer 2

Here is the code that I tried.

I think you can't ask for a simpler code than this.

    String s = "aaaabbbbffffd", modified = "";
    int len = s.length(),  i = 0,  j = i + 1,  count = 1;
    char[] c = s.toCharArray();
    for (; i < len; i = j) {
        count = 1;
            for (; j < len; j++) 
                 if (c[i] == c[j]) 
                 count++;
                    else {
                    j++;
                    break;
                    }
        modified += c[i] + "" + count;
    }
    System.out.println(modified);

Answer 3

Here is my solution

public String countChars(String in){
 LinkedHashMapMap<Character, Integer> map = new LinkedHashMap<Character, Integer>();
 for(char c: in.toCharArray()){
   Integer count =  map.get(c);
   if(count==null){
    count=0;
   }
   count++;
   map.put(c,count);
 }
 String out ="";
 for(Entry<Character, Integer> e : map.entrySet()){
    out += e.getKey()+e.getValue();
 }
 return out;
}

Answer 4

Employ a Map<Character, Integer> instead. Attempt to insert the new character into the map; if it already exists, then increment the value for that particular character.

Example:

Map<Character, Integer> countMap = new HashMap<>();
if(!countMap.containsKey('a')) {
    countMap.put('a', 1);
} else {
    countMap.put('a', countMap.get('a') + 1);
}

Answer 5

My version

    StringBuilder sb = new StringBuilder();
    int count = 0;
    char last = s.charAt(0);
    for(char c : s.toCharArray()) {
        if (c == last) {
            count++;
        } else {
            sb.append(last).append(count);
            count = 0;
            last = c;
        }
    }
    if (count != 0) {
        sb.append(last).append(count);
    }
    System.out.println(sb);