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(the following part is heavily inspired from my other answer to a recent, similar question)

edit: Once again, @Misha's answer is a welcome improvement of mine, do take a look at that too!

(the following part is heavily inspired from my other answer to a recent, similar question)

edit: Once again, @Misha's answer is a welcome improvement of mine, do take a look at that too!

edit: Once again, @Misha's answer is a welcome improvement of mine, do take a look at that too!

public int rarest(Map<String, Integer> m)throws Exception

1. Count the number of occurrences for each age

   • You can apply a groupingBy() on m's values, using Collectors.counting() to populate the intermediate map's values:

       // assume map is the method argument
       Map<Integer, Long> temp = map.entrySet().stream()
           .collect(Collectors.groupingBy(Entry::getValue, Collectors.counting()))
     

2. Sort by the least values, then by the youngest age

   • From this intermediate map, you can stream the entries, sort by the values using Entry.comparingByValue() and then return the first entry, i.e. the rarest:

       temp.entrySet().stream()
           .sorted(Entry.comparingByValue())
           .findFirst().map(Entry::getKey).get().intValue();
     

public static int rarest(Map<String, Integer> map) {
    return map.entrySet().stream()
            .collect(Collectors.groupingBy(Entry::getValue,
                    Collectors.counting()))
            .entrySet().stream()
            .sorted(Entry.comparingByValue())
            .findFirst().map(Entry::getKey).get().intValue();
}

public int rarest(Map<String, Integer> m) throws Exception

1. Count the number of occurrences for each age

   • You can apply a groupingBy() on m's values, using Collectors.counting() to populate the intermediate map's values:

       // assume map is the method argument
       Map<Integer, Long> temp = map.entrySet().stream()
           .collect(Collectors.groupingBy(Entry::getValue, Collectors.counting()))
     

2. Sort by the least values, then by the youngest age

   • From this intermediate map, you can stream the entries, sort by the values using a custom Comparator lambda declaration and then return the first entry, i.e. the rarest:

       temp.entrySet().stream()
           .sorted((e1, e2) -> { 
               int v = e1.getValue().compareTo(e2.getValue());
               return v != 0 ? v : e1.getKey().compareTo(e2.getKey());
           })
           .findFirst().map(Entry::getKey).get().intValue();
     

public static int rarest(Map<String, Integer> map) {
    return map.entrySet().stream()
            .collect(Collectors.groupingBy(Entry::getValue,
                    Collectors.counting()))
            .entrySet().stream()
            .sorted((e1, e2) -> { 
                int v = e1.getValue().compareTo(e2.getValue());
                return v != 0 ? v : e1.getKey().compareTo(e2.getKey());
            })
            .findFirst().map(Entry::getKey).get().intValue();
}