How to put numbers into an array and sorted by most frequent number in java - numbers

Question

I was given this question on programming in java and was wondering what would be the best way of doing it.
The question was on the lines of:
From the numbers provided, how would you in java display the most frequent number. The numbers was: 0, 3, 4, 1, 1, 3, 7, 9, 1
At first I am thinking well they should be in an array and sorted first then maybe have to go through a for loop. Am I on the right lines. Some examples will help greatly

Answer 1

If the numbers are all fairly small, you can quickly get the most frequent value by creating an array to keep track of the count for each number. The algorithm would be:
Find the maximum value in your list
Create an integer array of size max + 1 (assuming all non-negative values) to store the counts for each value in your list
Loop through your list and increment the count at the index of each value
Scan through the count array and find the index with the highest value
The run-time of this algorithm should be faster than sorting the list and finding the longest string of duplicate values. The tradeoff is that it takes up more memory if the values in your list are very large.

Answer 2

With Java 8, this can be implemented rather smoothly. If you're willing to use a third-party library like jOOλ, it could be done like this:
List<Integer> list = Arrays.asList(0, 3, 4, 1, 1, 3, 7, 9, 1);
System.out.println(
Seq.seq(list)
.grouped(i -> i, Agg.count())
.sorted(Comparator.comparing(t -> -t.v2))
.map(t -> t.v1)
.toList());
(disclaimer, I work for the company behind jOOλ)
If you want to stick with the JDK 8 dependency, the following code would be equivalent to the above:
System.out.println(
list.stream()
.collect(Collectors.groupingBy(i -> i, Collectors.counting()))
.entrySet()
.stream()
.sorted(Comparator.comparing(e -> -e.getValue()))
.map(e -> e.getKey())
.collect(Collectors.toList()));
Both solutions yield:
[1, 3, 0, 4, 7, 9]