Been a while since this question was asked but these days I'm partial to something like:

public static <K, V> Map<K, V> zipToMap(List<K> keys, List<V> values) {
    return IntStream.range(0, keys.size()).boxed()
            .collect(Collectors.toMap(keys::get, values::get));
}

For those unfamiliar with streams, what this does is gets an IntStream from 0 to the length, then boxes it, making it a Stream<Integer> so that it can be transformed into an object, then collects them using Collectors.toMap which takes two suppliers, one of which generates the keys, the other the values.

This could stand some validation (like requiring keys.size() be less than values.size()) but it works great as a simple solution.

EDIT: The above works great for anything with constant time lookup, but if you want something that will work on the same order (and still use this same sort of pattern) you could do something like:

public static <K, V> Map<K, V> zipToMap(List<K> keys, List<V> values) {
    Iterator<K> keyIter = keys.iterator();
    Iterator<V> valIter = values.iterator();
    return IntStream.range(0, keys.size()).boxed()
            .collect(Collectors.toMap(_i -> keyIter.next(), _i -> valIter.next()));
}

The output is the same (again, missing length checks, etc.) but the time complexity isn't dependent on the implementation of the get method for whatever list is used.

Assuming, that both lists are of equal length and that keys and values have the same index in both lists, then this is an easy approach:

for (int i = 0; i < strings.size(); i++) {
  map.put(strings.get(i), integers.get(i));
}

Instead of using an auxiliary list to hold the indices, you can have them generated by an IntStream.

Map<Double, String> map = IntStream.range(0, list1.size())
            .boxed()
            .collect(Collectors.toMap(i -> list1.get(i), i -> list2.get(i)));

You can do this by iterating over all the entries in mapAdded and merging them into mapGlobal.

The following iterates over the entries of mapAdded by calling forEach(action) where the action consumes the key and value of each entry. For each entry, we call merge(key, value, remappingFunction) on mapGlobal: this will either create the entry under the key k and value v if the key didn't exist or it will invoke the given remapping function if they already existed. This function takes the 2 lists to merge, which in this case, are first added to a TreeSet to ensure both unique and sorted elements and converted back into a list:

mapAdded.forEach((k, v) -> mapGlobal.merge(k, v, (v1, v2) -> {
    Set<String> set = new TreeSet<>(v1);
    set.addAll(v2);
    return new ArrayList<>(set);
}));

If you want to run that potentially in parallel, you can create a Stream pipeline by getting the entrySet() and calling parallelStream() on it. But then, you need to make sure to use a map that supports concurrency for mapGlobal, like a ConcurrentHashMap.

ConcurrentMap<String, List<String>> mapGlobal = new ConcurrentHashMap<>();
// ...
mapAdded.entrySet().parallelStream().forEach(e -> mapGlobal.merge(e.getKey(), e.getValue(), (v1, v2) -> {
    Set<String> set = new TreeSet<>(v1);
    set.addAll(v2);
    return new ArrayList<>(set);
}));

This does not look horribly wrong to me. There are some issues though:

1. groupingBy I presume is a static import from Collectors? You should make these things more obvious. In fact, I dislike the static imports for most things.....
2. you don't show much about ruleKeys.

The style otherwise looks good.

The concept you are solving, given the information you have given, is being solved in what amounts to a reasonably efficient manner, all things considered. It is an \$O(nm)\$ solution, if you double either the number of rules, or number of keys, your solution will take twice as long, and if you double both, the solution will take 4 times as long.

If you do this just once and then reuse the Mapping, then it should be fine. If you have fewer than say 100 members in each of the rules/keys, then you should be fine.

If you have more data than that then you should consider building up the data directly in to the map at the same time as you initially populate the rules/keys.

So, you are right to not like the solution because it scales badly, but, if your data is small, be pragmatic and not worry about it too much in this instance.

Easiest way is to use Map to group by operators.Actually you can use Map instead of List but you can also iterate through map to append key and values and insert them back to the list. You can use the following code to achieve the result you want.

public static void main(String[] args) {
    List<Object> list1 = new ArrayList<Object>();
    list1.add("operator1, id1");
    list1.add("operator1, id2");
    list1.add("operator2, id3");
    list1.add("operator2, id4");
    list1.add("operator3, id5");
    list1.add("operator4, id6");

    List<Object> list2 = new ArrayList<Object>();
    list2.add("operator4, id7");
    list2.add("operator3, id8");
    list2.add("operator2, id9");
    list2.add("operator2, id10");
    list2.add("operator1, id11");
    list2.add("operator1, id12");

    Map<String, String> map = new HashMap<String, String>();
    List<Object> list3 = new ArrayList<Object>();

    Iterator<Object> it1 = list1.iterator();
    Iterator<Object> it2 = list2.iterator();

    while (it1.hasNext() && it2.hasNext()) {
        String[] line1 = ((String) it1.next()).split(",");
        map.put(line1[0], map.get(line1[0]) == null ? line1[1] : map.get(line1[0]) + ", " + line1[1]);

        String[] line2 = ((String) it2.next()).split(",");
        map.put(line2[0], map.get(line2[0]) == null ? line2[1] : map.get(line2[0]) + ", " + line2[1]);
    }

    for (Map.Entry<String, String> entry : map.entrySet()) {
        String item = entry.getKey() + "," + entry.getValue();
        System.out.println(item);
        list3.add(item);
    }
}

After running the code above the result will be :

operator4, id6,  id7
operator1, id1,  id2,  id11,  id12
operator3, id5,  id8
operator2, id3,  id4,  id9,  id10

You can iterate over the indices of the Lists with an IntStream:

Map<String, String> result =
    IntStream.range(0,keys.size())
             .boxed()
             .collect(Collectors.toMap(i -> keys.get(i), i -> values.get(i)));

You should check if locations iterator has next element

if (locations.hasNext()) {
    tag = locations.next();
} else {
    tag = null;
}

final List<HashMap<String, String>> joinedById = list1.stream()
            .flatMap(m1 -> list2.stream()
                    .filter(y -> m1.get("id").equals(y.get("id")))
                    .map(m2 -> new HashMap<String, String>() {{
                        putAll(m1);
                        putAll(m2);
                    }}))
            .collect(Collectors.toList());

You can do something like this:

List<Food> foodList = fruits.keySet().stream()
    .concat(veggies.keySet().stream())
    .distinct()
    .map(id -> {
        Food food = new Food();
        food.setId(id);
        food.setFruitsList(fruits.getOrDefault(id, new ArrayList<>()));
        food.setVeggiesList(veggies.getOrDefault(id, new ArrayList<>()));
        return food ;
    })
    .collect(Collectors.toList());

If your Food class has a constructor taking the three parameters, it gets just a little more concise:

List<Food> foodList = fruits.keySet().stream()
    .concat(veggies.keySet().stream())
    .distinct()
    .map(id -> new Food(
        id,
        fruits.getOrDefault(id, new ArrayList<>()),
        veggies.getOrDefault(id, new ArrayList<>())
    )
    .collect(Collectors.toList());

If you know the ids are the same in both maps, you can skip the .concat() and .distinct() steps (which are quite expensive), and just use get() instead of getOrDefault().

You can also (in any case) do

Set<Integer> allKeys = new HashSet<>(fruits.keySet());
allKeys.addAll(veggies.keySet());
List<Food> foodList = allKeys.stream()
    .map(/* as before */)
    .collect(Collectors.toList());

which is not quite a "pure stream" a solution, but is probably more efficient.

This should be optimal. You first build a map from the currencies to their commercial banks. Then you run through your centrals building a map from commercial to central (looked up in the first map).

    List<CurrencyItem> currenciesByCommercialBank = new ArrayList<>();
    List<CurrencyItem> currenciesByCentralBank = new ArrayList<>();
    // Build my lookup from CurrencyName to CommercialBank.
    Map<CurrencyName, CurrencyItem> commercials = currenciesByCommercialBank
            .stream()
            .collect(
                    Collectors.toMap(
                            // Map from currency name.
                            ci -> ci.getName(),
                            // To the commercial bank itself.
                            ci -> ci));
    Map<CurrencyItem, CurrencyItem> commercialToCentral = currenciesByCentralBank
            .stream()
            .collect(
                    Collectors.toMap(
                            // Map from the equivalent commercial
                            ci -> commercials.get(ci.getName()),
                            // To this central.
                            ci -> ci
                    ));

This is my implementation, i generalized for n lists.

public static Map<String,Object> mixLists(List<?>...lists){
    Map<String,Object> map = new HashMap<String,Object>();

    for (int i = 0;i < lists.length;i++){
        if (!lists[i].isEmpty())
            map.put(lists[i].get(0).getClass().getName(), lists[i]);
    }
    return map;
}

The function returns a map that you can iterate like a List.