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Java Comparator Sorting Problem Using Map Values

I have a scenario where I need to take the keys of a Map<String, Set<String>>, and add them into a new Set<String> that is sorted. The sort order is based on the Map values for each key. The value for each key of the map is a Set containing other keys that are related to that key.
I need the keys to be sorted in such a way that a related key must be BEFORE another key that contains it in its related Set. To use a programming paradigm, it's similar to requiring a variable to be declared on an earlier line, before it can be referenced on another line.
For example, the following represents the contents of the Map<String, Set<String>>:
abc=[def, ghi, jkl, mno]
def=[]
ghi=[def]
jkl=[ghi, stu]
mno=[]
pqr=[abc]
stu=[def]
vwx=[mno, ghi]
zy0=[jkl]
In this example, the key "jkl" has a relationship to keys, "ghi" and "stu", "def" does not have a relationship to any of the keys.
NOTE: The relationships will be ONE-WAY only. So, for example, if "ghi" is related to "def", "def" will NEVER be related to "ghi".
So, for the above Map, the sort order would be:
def=[]
mno=[]
ghi=[def]
stu=[def]
vwx=[mno, ghi]
jkl=[ghi, stu]
zy0=[jkl]
abc=[def, ghi, jkl, mno]
pqr=[abc]
Here's the Comparator that I wrote. It's inside of a runnable test class that uses the example above:
import java.util.*;
public class RelationshipComparator_Test {
public static void main(String[] args) {
String[] testMap = "abc=[def,ghi,jkl,mno]|def=[]|ghi=[def]|jkl=[ghi,stu]|mno=[]|pqr=[abc]|stu=[def]|vwx=[mno,ghi]|zy0=[jkl]".split("[|]");
Map<String, Set<String>> relationshipMap = new HashMap<>();
for (String entry : testMap) {
String[] keyValue = entry.split("[=]");
String replacement = keyValue[1].replaceAll("[^a-z0-9,]", "");
Set<String> valueSet = new HashSet<>();
String[] values = (!replacement.equals("") ? replacement.split("[,]") : new String[0]);
Collections.addAll(valueSet, values);
relationshipMap.put(keyValue[0], valueSet);
}
Set<String> sortedKeys = new TreeSet<>(new RelationshipComparator(relationshipMap));
sortedKeys.addAll(relationshipMap.keySet());
for (String key : sortedKeys) {
System.out.println(key + "=" + relationshipMap.get(key));
}
}
static class RelationshipComparator implements Comparator<String> {
private Map<String, Set<String>> relationshipMap;
RelationshipComparator(Map<String, Set<String>> relationshipMap) {
this.relationshipMap = relationshipMap;
}
#Override
public int compare(String o1, String o2) {
Set<String> o1Set = relationshipMap.get(o1);
Set<String> o2Set = relationshipMap.get(o2);
if (o1Set != null && o2Set != null) {
if (o1Set.size() == 0 && o2Set.size() > 0) {
printCompare(o1, o2, "o1Set.size() == 0: -1");
return -1;
}
if (o2Set.size() == 0 && o1Set.size() > 0) {
printCompare(o1, o2, "o2Set.size() == 0: 1");
return 1;
}
if (o1Set.contains(o2)) {
printCompare(o1, o2, "o1Set.contains(o2): 1");
return 1;
}
if (o2Set.contains(o1)) {
printCompare(o1, o2, "o2Set.contains(o1): -1");
return -1;
}
}
printCompare(o1, o2, "default: " + o1.compareTo(o2));
return o1.compareTo(o2);
}
private void printCompare(String o1, String o2, String result) {
System.out.println("**********");
System.out.println("o1: " + o1 + "=" + relationshipMap.get(o1));
System.out.println("o2: " + o2 + "=" + relationshipMap.get(o2));
System.out.println("result: " + result);
System.out.println("**********");
System.out.println();
}
}
}
If you run the code, you'll see the following output:
def=[]
mno=[]
ghi=[def]
jkl=[stu, ghi]
abc=[def, ghi, jkl, mno]
pqr=[abc]
stu=[def]
vwx=[ghi, mno]
zy0=[jkl]
It's incorrect because, "jkl" references "stu", but "stu" is sorted after "jkl".
Any help would be greatly appreciated.

You say that relationships are one-way, which rules out obvious cases such as:
a=[b]
b=[a]
for which no solution is possible. However, we also need to rule out cyclic relationships such as:
a=[b]
b=[c]
c=[a]
If this is the case then I believe you can achieve the required ordering by using a PriorityQueue to order keys by the size of the value set related to the key. As keys are removed from the queue they also have to be removed from any of the related value sets that contain them. Which value sets contain a given key can be recovered from a reverse Map<String, Set<String>> which holds the set of keys that refer to a given value key.
Hopefully some code will make things clearer:
static List<String> orderByRef(Map<String, Set<String>> relationshipMap)
{
final Map<String, Set<String>> relationshipMapCopy = new HashMap<>();
for(String key : relationshipMap.keySet())
relationshipMapCopy.put(key, new HashSet<>(relationshipMap.get(key)));
final Map<String, Set<String>> referencedBy = new HashMap<>();
for(String key : relationshipMap.keySet())
referencedBy.put(key, new HashSet<>());
for (Entry<String,Set<String>> e : relationshipMapCopy.entrySet())
for(String v : e.getValue())
referencedBy.get(v).add(e.getKey());
PriorityQueue<String> pq = new PriorityQueue<>(new Comparator<String>()
{
#Override
public int compare(String k1, String k2)
{
return relationshipMapCopy.get(k1).size() - relationshipMapCopy.get(k2).size();
}
});
pq.addAll(relationshipMap.keySet());
List<String> orderedKeys = new ArrayList<>();
while(!pq.isEmpty())
{
String minKey = pq.poll();
if(!relationshipMapCopy.get(minKey).isEmpty())
{
// cyclic relationship
break;
}
orderedKeys.add(minKey);
for(String refKey : referencedBy.get(minKey))
{
// remove minKey from value set of refKey
relationshipMapCopy.get(refKey).remove(minKey);
// reorder refKey in pq
pq.remove(refKey);
pq.add(refKey);
}
}
return orderedKeys;
}
Note that since we're modifying the relationshipMap by removing keys from value sets we first need to create a deep copy. Also, we can detect the presence of a cyclic relationships by checking that the value set of the min key is empty.
Output:
def []
mno []
stu [def]
ghi [def]
vwx [ghi, mno]
jkl [stu, ghi]
zy0 [jkl]
abc [def, ghi, jkl, mno]
pqr [abc]
Which satisfies the constraint that no key is referenced before it appears in the list.
For input containing a cyclic relationship, eg (z=[y]|y=[]|a=[b]|b=[c]|c=[a]), we get:
y []
z [y]

How to count occurrences for each value in MultiMap ? (java)

I have Multimap, which contains two strings. Example:
1 = [key,car],
2 = [key,blue],
3 = [key,car]
Multimap definition (I am using Guava library):
ListMultimap<Integer, String> map_multi = ArrayListMultimap.create();
And this is how I put values in MultiMap:
for (int i = 0; i < list.size(); i++) {
if (i + 1 < list.size()) {
multimap.put(i,(String) list.get(i));
multimap.put(i,(String) list.get(i+1));
} else if (i + 1 == list.size()) {
}
}
I want to count the occurrences of the same value inside the multimap.
So the result should be 2 if i count how many values [key,car] are (per example I have given above) in my multimap:
occurrences of [key,car] = 2
occurrences of [key,blue] = 1
I have also tried to implement this with multi value HashMap and I was counting it with this, way (Storage is class where I store two string values inside object):
B = Collections.frequency(new ArrayList<Storage>(map.values()), map.get(number));
But I don't get the right results.

You can achieve what you want by creating a map that has your multimap values as the keys and the count as the value:
Map<Collection<String>, Long> result = map_multi.asMap().values().stream()
.collect(Collectors.groupingBy(v -> v, Collectors.counting()));
Here I've used Guava's Multimap.asMap method to get a view over the original multimap, then collected the values into a new map.
Another way, without streams:
Map<Collection<String>, Integer> result = new HashMap<>();
map_multi.asMap().values().forEach(v -> result.merge(v, 1, Integer::sum));
This uses the Map.merge method to accumulate equal values by counting its occurrences.

Please try this code. map_multi.get(key).size() is your answer.
ListMultimap<Integer, String> map_multi = ArrayListMultimap.create();
map_multi.put(1, "car");
map_multi.put(2, "blue");
map_multi.put(3, "apple");
map_multi.put(1, "car");
for (Integer key : map_multi.keySet()) {
System.out.println(map_multi.get(key).get(0) + " occurances: " + map_multi.get(key).size());
}
Output:
car occurances: 2
blue occurances: 1
apple occurances: 1

So the first step is to create a Map<Integer, List<String>> from your ListMultimap. You can do this in the following way:
Map<Integer, List<String>> collect = map_multi.entries()
.stream()
.collect(Collectors.groupingBy(Map.Entry::getKey,
Collectors.mapping(Map.Entry::getValue,
Collectors.toList())));
Then let's say you have a List<String> with car and key in it.
Example:
List<String> myList = List.of("key", "car"); // java 9
You just iterate through the values() of the map and check if myList contains all elements from map's lists.
long count = collect.values()
.stream()
.filter(list -> list.containsAll(myList))
.count();

I think you're using wrong collection to store your data. Based on what you wrote, you want a map with integer keys and two-element tuples as values and then use Multiset to count frequencies:
Multiset is a collection that supports order-independent equality, like Set, but may have duplicate elements. A multiset is also sometimes called a bag.
Elements of a multiset that are equal to one another are referred to as occurrences of the same single element. The total number of occurrences of an element in a multiset is called the count of that element (the terms "frequency" and "multiplicity" are equivalent, but not used in this API).
The code below assumes you have proper implementation of two-element tuple (aka pair, or your Storage class, but with proper equals and hashCode implementations), like this one from jOOL:
HashMap<Integer, Tuple2<String, String>> m = new HashMap<>();
Tuple2<String, String> carTuple = new Tuple2<>("key", "car");
Tuple2<String, String> blueTuple = new Tuple2<>("key", "blue");
m.put(1, carTuple);
m.put(2, blueTuple);
m.put(3, carTuple);
ImmutableMultiset<Tuple2<String, String>> occurrences =
ImmutableMultiset.copyOf(m.values());
System.out.println(occurrences); // [(key, car) x 2, (key, blue)]
If you need to have few values (tuples) mapped under one key (integer), then you should change the first line to multimap:
ListMultimap<Integer, Tuple2<String, String>> m = ArrayListMultimap.create();
so that m.put(1, anotherTuple) is possible and putting doesn't override first value (carTuple) but rather adds it under 1 values list.
EDIT:
You can implement Tuple2 yourself if you don't need/want additional dependency, it could look like this class:
public class Tuple2<T1, T2> {
public final T1 v1;
public final T2 v2;
public Tuple2(T1 v1, T2 v2) {
this.v1 = v1;
this.v2 = v2;
}
#Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (!(o instanceof Tuple2)) {
return false;
}
#SuppressWarnings({"unchecked", "rawtypes"}) final Tuple2<T1, T2> that = (Tuple2) o;
return Objects.equals(v1, that.v1) && Objects.equals(v2, that.v2);
}
#Override
public int hashCode() {
return Objects.hash(v1, v2);
}
#Override
public String toString() {
return "(" + v1 + ", " + v2 + ")";
}
}

How to sort " ArrayList<HashMap<String, String>> arrList " alphabatically?

Can you guide me how can I sort Array List having Hash Map alphabatically?
JSONArray jArr2 = new JSONArray(jsonString2);
for(int i=0;i<jArr2.length();i++){
HashMap<String, String> map = new HashMap<String, String>();
map.put("titleName",jArr2.getJSONObject(i).getString("titleName"))
programList.add(map);
}

Implement a Comparator<HashMap<String, String>> which just extracts the value assocated with the value key, then use Collections.sort method to sort your arraylist.
For e.g.:
class MyComparator implements Comparator<Map<String, String>>{
private final String key;
public MyComparator(String key)
{
this.key = key;
}
public int compare(Map<String, String> first,
Map<String, String> second)
{
// TODO: Null checking, both for maps and values
String firstValue = first.get(key);
String secondValue = second.get(key);
return firstValue.compareTo(secondValue);
}
}

Looking at your example, I don't think you need a Map to be involved at all. What you have is a list of Maps, where every Map only has one key, which is "titleName". Why not just have a list of titlenames? Then your code would look like this:
JSONArray jArr2 = new JSONArray(jsonString2);
List<String> titleNames = new ArrayList<>();
for (int i = 0; i < jArr2.length(); i++) {
titleNames.add(jArr2.getJSONObject(i).getString("titleName"))
}
You know that the list only contains titleNames, you don't need to complicate the data structure with Maps!
Then you can sort the list simply by using
Collections.sort(titleNames);
Note that this will work while the other answers that suggests Collections.sort() on the list of maps will not work. This is because titleNamees is a List of Strings, which implement Comparable (ie the sort() method knows how to order them with respect to each other), while Map does not implement comparable (as there are multiple ways to order Maps - number of entries, total number of bytes, etc).

Yes you can use Collections.sort(); with a custom comparator. Here is the doc.
Collections.sort(YOUR_ARRAY_LIST, new YourCustomComparator());
And this should be the class you must have
class YourCustomComparator implements Comparator<HashMap<String, String>> {
#Override
public int compare(HashMap<String, String> lhs, HashMap<String, String> rhs) {
// check here your objects. lhs and rhs. compare them as you want
// return 1 if lhs is greater than rhs
// return 0 if ther are same
// return -1 otherwise
}
}

Java convert {String,String}[] to Map<String,String[]>

Given the class:
public class CategoryValuePair
{
String category;
String value;
}
And a method:
public Map<String,List<String>> convert(CategoryValuePair[] values);
Given that in values we can receive many entries with the same category, I want to convert these into a Map grouped on category.
Is there a quick / efficient way to perform this conversion?

As far as I know there is not easier way than iterating on values, and then putting the values in the map (like some predefined method).
Map<String, List<String>> map = new HashMap<String, List<String>>();
if (values != null) {
for (CategoryValuePair cvp : values) {
List<String> vals = map.get(cvp.category);
if (vals == null) {
vals = new ArrayList<String>();
map.put(cvp.category, vals);
}
vals.add(cvp.value);
}
}
I changed the map values from String[] to List<String> since it seems easier to me to use that so you don't have to hassle with array resizing.

To make it in fewer lines of code, use Google Collections:
public Map<String, Collection<String>> convert(CategoryValuePair[] values) {
Multimap<String, String> mmap = ArrayListMultimap.create();
for (CategoryValuePair value : values) {
mmap.put(value.category, value.value);
}
return mmap.asMap();
}
If you don't want to allow duplicate values, replace ArrayListMultimap with HashMultimap.

With lambdaj you just need one line of code to achieve that result as it follows:
group(values, by(on(CategoryValuePair.class).getCategory()));

Just for the sake of implementation... The method returns Map and also checks for duplicates in the arrays... though performance wise its heavy ...
public Map<String,String[]> convert(CategoryValuePair[] values)
{
Map<String, String[]> map = new HashMap<String, String[]>();
for (int i = 0; i < values.length; i++) {
if(map.containsKey(values[i].category)){
Set<String> set = new HashSet<String>(Arrays.asList(map.get(values[i].category)));
set.add(values[i].value);
map.put(values[i].category, set.toArray(new String[set.size()]));
}else {
map.put(values[i].category, new String[]{values[i].value});
}
}
return map;
}

TreeMap sort by value

I want to write a comparator that will let me sort a TreeMap by value instead of the default natural ordering.
I tried something like this, but can't find out what went wrong:
import java.util.*;
class treeMap {
public static void main(String[] args) {
System.out.println("the main");
byValue cmp = new byValue();
Map<String, Integer> map = new TreeMap<String, Integer>(cmp);
map.put("de",10);
map.put("ab", 20);
map.put("a",5);
for (Map.Entry<String,Integer> pair: map.entrySet()) {
System.out.println(pair.getKey()+":"+pair.getValue());
}
}
}
class byValue implements Comparator<Map.Entry<String,Integer>> {
public int compare(Map.Entry<String,Integer> e1, Map.Entry<String,Integer> e2) {
if (e1.getValue() < e2.getValue()){
return 1;
} else if (e1.getValue() == e2.getValue()) {
return 0;
} else {
return -1;
}
}
}
I guess what am I asking is: Can I get a Map.Entry passed to the comparator?

You can't have the TreeMap itself sort on the values, since that defies the SortedMap specification:
A Map that further provides a total ordering on its keys.
However, using an external collection, you can always sort Map.entrySet() however you wish, either by keys, values, or even a combination(!!) of the two.
Here's a generic method that returns a SortedSet of Map.Entry, given a Map whose values are Comparable:
static <K,V extends Comparable<? super V>>
SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
return res != 0 ? res : 1;
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
Now you can do the following:
Map<String,Integer> map = new TreeMap<String,Integer>();
map.put("A", 3);
map.put("B", 2);
map.put("C", 1);
System.out.println(map);
// prints "{A=3, B=2, C=1}"
System.out.println(entriesSortedByValues(map));
// prints "[C=1, B=2, A=3]"
Note that funky stuff will happen if you try to modify either the SortedSet itself, or the Map.Entry within, because this is no longer a "view" of the original map like entrySet() is.
Generally speaking, the need to sort a map's entries by its values is atypical.
Note on == for Integer
Your original comparator compares Integer using ==. This is almost always wrong, since == with Integer operands is a reference equality, not value equality.
System.out.println(new Integer(0) == new Integer(0)); // prints "false"!!!
Related questions
When comparing two Integers in Java does auto-unboxing occur? (NO!!!)
Is it guaranteed that new Integer(i) == i in Java? (YES!!!)

polygenelubricants answer is almost perfect. It has one important bug though. It will not handle map entries where the values are the same.
This code:...
Map<String, Integer> nonSortedMap = new HashMap<String, Integer>();
nonSortedMap.put("ape", 1);
nonSortedMap.put("pig", 3);
nonSortedMap.put("cow", 1);
nonSortedMap.put("frog", 2);
for (Entry<String, Integer> entry : entriesSortedByValues(nonSortedMap)) {
System.out.println(entry.getKey()+":"+entry.getValue());
}
Would output:
ape:1
frog:2
pig:3
Note how our cow dissapeared as it shared the value "1" with our ape :O!
This modification of the code solves that issue:
static <K,V extends Comparable<? super V>> SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
return res != 0 ? res : 1; // Special fix to preserve items with equal values
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}

In Java 8:
LinkedHashMap<Integer, String> sortedMap = map.entrySet().stream()
.sorted(Map.Entry.comparingByValue(/* Optional: Comparator.reverseOrder() */))
.collect(Collectors.toMap(Map.Entry::getKey,
Map.Entry::getValue,
(e1, e2) -> e1, LinkedHashMap::new));

A TreeMap is always sorted by the keys, anything else is impossible. A Comparator merely allows you to control how the keys are sorted.
If you want the sorted values, you have to extract them into a List and sort that.

This can't be done by using a Comparator, as it will always get the key of the map to compare. TreeMap can only sort by the key.

Olof's answer is good, but it needs one more thing before it's perfect. In the comments below his answer, dacwe (correctly) points out that his implementation violates the Compare/Equals contract for Sets. If you try to call contains or remove on an entry that's clearly in the set, the set won't recognize it because of the code that allows entries with equal values to be placed in the set. So, in order to fix this, we need to test for equality between the keys:
static <K,V extends Comparable<? super V>> SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
if (e1.getKey().equals(e2.getKey())) {
return res; // Code will now handle equality properly
} else {
return res != 0 ? res : 1; // While still adding all entries
}
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
"Note that the ordering maintained by a sorted set (whether or not an explicit comparator is provided) must be consistent with equals if the sorted set is to correctly implement the Set interface... the Set interface is defined in terms of the equals operation, but a sorted set performs all element comparisons using its compareTo (or compare) method, so two elements that are deemed equal by this method are, from the standpoint of the sorted set, equal."
(http://docs.oracle.com/javase/6/docs/api/java/util/SortedSet.html)
Since we originally overlooked equality in order to force the set to add equal valued entries, now we have to test for equality in the keys in order for the set to actually return the entry you're looking for. This is kinda messy and definitely not how sets were intended to be used - but it works.

I know this post specifically asks for sorting a TreeMap by values, but for those of us that don't really care about implementation but do want a solution that keeps the collection sorted as elements are added, I would appreciate feedback on this TreeSet-based solution. For one, elements are not easily retrieved by key, but for the use case I had at hand (finding the n keys with the lowest values), this was not a requirement.
TreeSet<Map.Entry<Integer, Double>> set = new TreeSet<>(new Comparator<Map.Entry<Integer, Double>>()
{
#Override
public int compare(Map.Entry<Integer, Double> o1, Map.Entry<Integer, Double> o2)
{
int valueComparison = o1.getValue().compareTo(o2.getValue());
return valueComparison == 0 ? o1.getKey().compareTo(o2.getKey()) : valueComparison;
}
});
int key = 5;
double value = 1.0;
set.add(new AbstractMap.SimpleEntry<>(key, value));

A lot of people hear adviced to use List and i prefer to use it as well
here are two methods you need to sort the entries of the Map according to their values.
static final Comparator<Entry<?, Double>> DOUBLE_VALUE_COMPARATOR =
new Comparator<Entry<?, Double>>() {
#Override
public int compare(Entry<?, Double> o1, Entry<?, Double> o2) {
return o1.getValue().compareTo(o2.getValue());
}
};
static final List<Entry<?, Double>> sortHashMapByDoubleValue(HashMap temp)
{
Set<Entry<?, Double>> entryOfMap = temp.entrySet();
List<Entry<?, Double>> entries = new ArrayList<Entry<?, Double>>(entryOfMap);
Collections.sort(entries, DOUBLE_VALUE_COMPARATOR);
return entries;
}

import java.util.*;
public class Main {
public static void main(String[] args) {
TreeMap<String, Integer> initTree = new TreeMap();
initTree.put("D", 0);
initTree.put("C", -3);
initTree.put("A", 43);
initTree.put("B", 32);
System.out.println("Sorted by keys:");
System.out.println(initTree);
List list = new ArrayList(initTree.entrySet());
Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {
#Override
public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2) {
return e1.getValue().compareTo(e2.getValue());
}
});
System.out.println("Sorted by values:");
System.out.println(list);
}
}

//convert HashMap into List
List<Entry<String, Integer>> list = new LinkedList<Entry<String, Integer>>(map.entrySet());
Collections.sort(list, (o1, o2) -> o1.getValue().compareTo(o2.getValue()));

If you want to use a Hash map you can add a condition in Comparator to check by values first & if values are equal perform a sort on keys
HashMap<String , Integer> polpularity = new HashMap<>();
List<String> collect = popularity.entrySet().stream().sorted((t2, t1) -> {
if (t2.getValue() > t1.getValue()) {
return -1;
} else if (t2.getValue() < t1.getValue()) {
return +1;
} else {
return t2.getKey().compareTo(t1.getKey());
}
}).map(entry -> entry.getKey()).collect(Collectors.toList());
If you don't want to take care of the latter condition then use a Treemap which will offer you sorting by itself, this can be done in an elegant single line of code:
TreeMap<String, Integer> popularity = new TreeMap<>();
List<String> collect = popularity.entrySet().stream().sorted(Collections.reverseOrder(Map.Entry.comparingByValue())).map(entry -> entry.getKey()).collect(Collectors.toList());

TreeMap is always sorted by the keys.
If you want TreeMap to be sorted by the values, so you can simply construct it also.
Example:
// the original TreeMap which is sorted by key
Map<String, Integer> map = new TreeMap<>();
map.put("de",10);
map.put("ab", 20);
map.put("a",5);
// expected output:
// {a=5, ab=20, de=10}
System.out.println(map);
// now we will constrcut a new TreeSet which is sorted by values
// [original TreeMap values will be the keys for this new TreeMap]
TreeMap<Integer, String> newTreeMapSortedByValue = new TreeMap();
treeMapmap.forEach((k,v) -> newTreeMapSortedByValue.put(v, k));
// expected output:
// {5=a, 10=de, 20=ab}
System.out.println(newTreeMapSortedByValue);

Only 1 Line Of Code Solution
Normal Order
map.entrySet().stream().sorted(Map.Entry.comparingByValue()).forEach(x->{});
Reverse Order
map.entrySet().stream().sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).forEachOrdered(x -> {});