CODEWITHSUNDEEP
javacollections
osimer pothe
osimer pothe

Reputation: 2897

Printing output of Collection.Max

I have the following code . Collections.Max returns the . How can I show the value of string and integer no By System.out.println() ? 

public class SortMapOnKeyExample {

    public static void main(String[] args) {
        List<String> list=new ArrayList<String>();
        list.add("sultan");
        list.add("Masum");
        list.add("sultan");
        list.add("Sorry");
        list.add("sultan");
        list.add("Masum");
        list.add("sultan");
        list.add("Tarek");
        list.add("sultan");

        Set<String> uniques = new HashSet(list);
        Map<String, Integer> counts = new HashMap<String, Integer>();

        for (String elem : uniques)
        {
            counts.put(elem, Collections.frequency(list, elem));
        }

        Collections.max(counts.entrySet(), new Comparator<Entry<String, Integer>>() {
            @Override
            public int compare(Entry<String, Integer> o1, Entry<String, Integer> o2) {
                return (o1.getValue() - o2.getValue());
            }
        });

    }

}

I have tried a lot But could not find out how can I do this ?please help me . The purpose of this code is to find the string that is occurred maximum and also its index ?

Upvotes: 0

Views: 160

Answers (1)

Boris the Spider
Boris the Spider

Reputation: 61168

Firstly, you code will run in O(n^2) time - each call to Collections.frequency must loop over the entire data, and this is done once for every element. You can easily make this O(n):

final Map<String, Integer> counts = new HashMap<>();
for (final String s : list) {
    final Integer c = counts.get(s);
    if (c == null) {
        counts.put(s, 1);
    } else {
        counts.put(s, c + 1);
    }
}

Now note that you can have more than one item with the same count. You need to sort the entries by value and then print the top ones:

final List<Entry<String, Integer>> entries = new ArrayList<>(counts.entrySet());
Collections.sort(entries, new Comparator<Entry<String, Integer>>() {

    @Override
    public int compare(final Entry<String, Integer> o1, final Entry<String, Integer> o2) {
        return Integer.compare(o2.getValue(), o1.getValue());
    }
});
final MessageFormat format = new MessageFormat("{0} has a count of {1,number,integer}");
final Iterator<Entry<String, Integer>> iter = entries.iterator();
final Entry<String, Integer> first = iter.next();
System.out.println(format.format(new Object[]{first.getKey(), first.getValue()}));
while (iter.hasNext()) {
    final Entry<String, Integer> entry = iter.next();
    if (entry.getValue() != first.getValue()) {
        break;
    }
    System.out.println(format.format(new Object[]{entry.getKey(), entry.getValue()}));
}

First we create a List from the entrySet() of the Map. Next we sort the List - notice the reversed order of the comparison - this means the sort is in descending rather than ascending order. Also note the use of Integer.compare, this is because using a - b to compare is very bad practice as it will overflow if a is large and b is large and negative; while not a problem here it is not a good habit to get into.

Now we take as Iterator of the List and keep printing out elements until we encounter one that is not equal to (must be less than) the count of the first element.

Output:

{sultan=5, Sorry=1, Tarek=1, Masum=2}
sultan has a count of 5

With different data, where we add Test five times also the output becomes:

{Test=5, sultan=5, Sorry=1, Tarek=1, Masum=2}
Test has a count of 5
sultan has a count of 5

Upvotes: 1

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