I'm trying to get a head start on practicing interview questions and I came across this one:
Turn String aaaabbbbddd into a4b4d3
You would basically want to convert the existing string into a string with each unique character occurrence and the number of times the character occurs.
This is my solution but I think it could be refined into something more elegant:
String s = "aaaabbbbddd";
String modified = "";
int len = s.length();
char[] c = s.toCharArray();
int count = 0;
for (int i = 0; i < len; i++) {
count = 1;
for (int j = i + 1; j < len; j++) {
if (c[i] == ' ') {
break;
}
if (c[i] == c[j]) {
count++;
c[j] = ' ';
}
}
if (c[i] != ' ') {
modified += c[i] + "" + count;
}
}
System.out.println(modified);
Does anyone have any other suggestions for a solution?
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);
}
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 = "aaaabbbbddd";
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
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);
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