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Collection Frequency in Java

beginner· Collections · Collections Utility

Problem

Collections.frequency() answers a narrower question than a full frequency map does — it counts occurrences of exactly one given value, without tallying every distinct value in the collection.

Given a List of integers, count how many times one specific value appears in it.

Input

[4, 7, 4, 2, 4, 9, 7], target = 4

Output

Frequency of 4: 3

Java Program

Java

import java.util.Arrays;
import java.util.Collections;
import java.util.List;

public class CollectionFrequency {
    public static void main(String[] args) {
        List<Integer> numbers = Arrays.asList(4, 7, 4, 2, 4, 9, 7);
        int count = Collections.frequency(numbers, 4); // counts only how many elements equal 4
        System.out.println("Frequency of 4: " + count);
    }
}

▶ Run Code

Output

Frequency of 4: 3

Core Logic

Passing the list and the target value to Collections.frequency() counts matching elements in one call, instead of a manual loop with its own counter.

How It Works

1Arrays.asList(4, 7, 4, 2, 4, 9, 7) builds the list of numbers to search.
2Collections.frequency(numbers, 4) compares every element against 4 using equals(), counting how many match.
3The method returns a plain int count — 3 here, since 4 appears three times.
4Nothing about the rest of the list's values matters — frequency() only reports on the one target value it was asked about.

In [4, 7, 4, 2, 4, 9, 7], the value 4 appears at three positions, so Collections.frequency(numbers, 4) returns 3.

💡

Key Point: This is a different task from building a full frequency map of every distinct value — frequency() is the right tool when only one specific value's count is actually needed, since it skips tracking the rest entirely.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: Collections.frequency() makes a single linear scan comparing every element to the target value, using only a running counter regardless of the list's contents.

Key Concepts

Collections.frequency()Listutility method

Approach 2: Java 8

Java

import java.util.Arrays;
import java.util.List;

public class CollectionFrequencyStream {
    public static void main(String[] args) {
        List<Integer> numbers = Arrays.asList(4, 7, 4, 2, 4, 9, 7);
        // Keeps only the matching elements, then tallies how many survived
        long count = numbers.stream().filter(n -> n == 4).count();
        System.out.println("Frequency of 4: " + count);
    }
}

▶ Run Code

Output

Frequency of 4: 3

Core Logic

Filtering the stream down to only the matching elements, then counting what's left, expresses the same 'how many equal this value' question as a pipeline.

How It Works

1numbers.stream() opens a stream over the list's elements.
2.filter(n -> n == 4) keeps only the elements equal to the target value, discarding everything else.
3.count() reduces the filtered stream down to how many elements survived — a long, not an int.
4No intermediate list of the matches is ever built — count() tallies as it filters.

Filtering [4, 7, 4, 2, 4, 9, 7] down to just the 4s keeps three elements, so count() returns 3.

💡

Key Point: count() returns a long rather than an int — a minor but real difference from Collections.frequency()'s plain int return, worth knowing if the result feeds into further arithmetic.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: filter() and count() together still make a single pass over the n elements, tallying matches without collecting them into a separate structure.

Key Concepts

Streamfilter()count()

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