Write a method rarest that accepts a map whose keys are strings and whose values are integers as a parameter and returns the integer value that occurs the fewest times in the map. If there is a tie, return the smaller integer value. If the map is empty, throw an exception.
For example, suppose the map contains mappings from students' names (strings) to their ages (integers). Your method would return the least frequently occurring age. Consider a map variable m containing the following key/value pairs:
{Alyssa=22, Char=25, Dan=25, Jeff=20, Kasey=20, Kim=20, Mogran=25, Ryan=25, Stef=22}
Three people are age 20 (Jeff, Kasey, and Kim), two people are age 22 (Alyssa and Stef), and four people are age 25 (Char, Dan, Mogran, and Ryan).
So a call of rarest(m) returns 22 because only two people are that age. If there is a tie (two or more rarest ages that occur the same number of times), return the youngest age among them. For example, if we added another pair of Kelly=22 to the map above, there would now be a tie of three people of age 20 (Jeff, Kasey, Kim) and three people of age 22 (Alyssa, Kelly, Stef). So a call of rarest(m) would now return 20 because 20 is the smaller of the rarest values.
Here is the link to the question.
public int rarest(Map<String, Integer> m)throws Exception{
if(m.size() == 0){
throw new Exception();//throwing an exception as per the question
}else{
Integer count = null;
//Creating a TreeMap to get the lowest age.
TreeMap<Integer, Integer> t = new TreeMap<Integer, Integer>();
for(Map.Entry<String, Integer> me : m.entrySet()){
if(t.containsKey(me.getValue())) {
count = t.get(me.getValue());
t.put(me.getValue(), count+1);
}else {
count = 1;
t.put(me.getValue(), count);
}
}
//If there is tie I am comparing the frequencies
int freq = t.get(t.firstKey());
TreeSet<Integer> val = new TreeSet<Integer>();
for(Integer i : t.keySet()){
if(freq > t.get(i) ){
freq = t.get(i);
val.add(i);
}
}
if(val.size() > 0){
return val.first();
}else{
return t.firstKey();
}
}
}
It gives the correct results for the given test cases, but I feel there should be a better way to do this.