I have an array of Strings:
String[] stringArray = {"x", "y", "z", "x", "x", "y", "a"};
What is the quickest/most efficient way to order this into a smaller Collection
in order of how frequent each String
is with its frequency?
I though about using the String
as a key in a HashMap<String,Integer>
but this wouldnt be sorted in terms of frequency
My other method i considered is using a TreeMap<Integer, String[]>
with a list of Strings with that integer, but there seems a lot of checking involved..
Im trying to avoid using more than one loop If possible as my String
arrays will be much larger than the one above. Thanks!
EDIT What i want is just to be able to output the Strings in order of frequency and preferably be able to pair that String with its frequency in the array, So for example two output arrays:
["x", "y", "z", "a"]
[3,2,1,1]
This would be quite a simple problem if speed wasnt an issue which is why i ask the great minds on here :)
Given a string str, the task is to sort the string according to the frequency of each character, in ascending order. If two elements have the same frequency, then they are sorted in lexicographical order. f, o, r occurs one time so they are ordered lexicographically and so are g, k and s.
To sort an array of strings in Java, we can use Arrays. sort() function.
The frequency sort algorithm is used to output elements of an array in descending order of their frequencies. If two elements have the same frequencies, then the element that occurs first in the input is printed first.
You can solve this in two steps:
Create a counter object - a Map<String, Integer>
listing for each string the number of times it appears in the input: in other words, it's a frequency map. This is O(n)
, as you only need to traverse the input once for building the map
With the previous map, create a list with its keys, sorted using the frequency of items (the values in the map) as ordering criteria. This is O(n log n)
, and you can call Collections.sort()
, with a Comparator
that uses the string frequency for the comparisons
This is what I mean:
String[] stringArray = {"x", "y", "z", "x", "x", "y", "a"};
final Map<String, Integer> counter = new HashMap<String, Integer>();
for (String str : stringArray)
counter.put(str, 1 + (counter.containsKey(str) ? counter.get(str) : 0));
List<String> list = new ArrayList<String>(counter.keySet());
Collections.sort(list, new Comparator<String>() {
@Override
public int compare(String x, String y) {
return counter.get(y) - counter.get(x);
}
});
After the above code executes, the variable list
will contain the following values (the order between elements of the same frequency is unspecified):
[x, y, a, z]
It's trivial to convert the list to an array:
list.toArray(new String[list.size()])
And if you need to find out the frequency of each string, just iterate over the sorted keys:
for (String str : list) {
int frequency = counter.get(str);
System.out.print(str + ":" + frequency + ", ");
}
Use the HashMap<String,Integer>
to maintain your counts. This will be the most efficient way to process the arbitrary list of strings.
Create an ArrayList<Map.Entry<String,Integer>>
from the map's entrySet()
.
Sort this list using a Collections.sort()
and a custom comparator.
Don't get hung up on micro-optimizations.
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