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Find Most Frequent Character in a String in Java

intermediate· Strings · String

Problem

The most frequent character is whichever character has the highest count once every character's occurrences have been tallied.

Given a string, find the character that appears the most times, along with its count.

Input

success

Output

Most frequent character: s (3 times)

Java Program

Java

import java.util.LinkedHashMap;
import java.util.Map;

public class MostFrequentCharacter {
    public static void main(String[] args) {
        String str = "success";
        Map<Character, Integer> freq = new LinkedHashMap<>();

        for (char c : str.toCharArray()) {
            freq.put(c, freq.getOrDefault(c, 0) + 1); // increment this character's running count
        }

        char maxChar = ' ';
        int maxCount = 0;
        for (Map.Entry<Character, Integer> entry : freq.entrySet()) {
            if (entry.getValue() > maxCount) { // found a new highest count so far
                maxChar = entry.getKey();
                maxCount = entry.getValue();
            }
        }

        System.out.println("Most frequent character: " + maxChar + " (" + maxCount + " times)");
    }
}

▶ Run Code

Output

Most frequent character: s (3 times)

Core Logic

Building the frequency map first, then scanning it once for the highest count, finds the winner in two simple passes.

How It Works

1A LinkedHashMap<Character, Integer> named freq tracks each character's count, exactly like counting character frequency.
2freq.getOrDefault(c, 0) reads the current count, or 0 if unseen, then freq.put(c, ...) stores the incremented count.
3maxChar and maxCount track the best candidate seen so far while walking entrySet().
4Whenever an entry's count exceeds maxCount, both maxChar and maxCount are updated to that entry.

In "success", 's' ends with a count of 3 — higher than 'u', 'c', or 'e' — so it's reported as the most frequent character.

💡

Key Point: Using > rather than >= when updating the maximum means the first character to reach the highest count wins any tie, keeping the result deterministic.

Complexity

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

Why: Building the frequency map visits every character once and holds one entry per distinct character, up to n in the worst case.

Key Concepts

LinkedHashMapgetOrDefault()running maximum

Approach 2: Java 8

Java

import java.util.Map;
import java.util.stream.Collectors;

public class MostFrequentCharacterStream {
    public static void main(String[] args) {
        String str = "success";

        Map<Character, Long> freq = str.chars()
            .mapToObj(c -> (char) c)
            .collect(Collectors.groupingBy(c -> c, Collectors.counting()));

        // Compares every entry by its count and keeps the largest one
        Map.Entry<Character, Long> maxEntry = freq.entrySet().stream()
            .max(Map.Entry.comparingByValue())
            .orElseThrow();

        System.out.println("Most frequent character: " + maxEntry.getKey() + " (" + maxEntry.getValue() + " times)");
    }
}

▶ Run Code

Output

Most frequent character: s (3 times)

Core Logic

Once the frequency map exists, a stream can find its highest-valued entry directly instead of a manual running-maximum loop.

How It Works

1str.chars().mapToObj(c -> (char) c) turns the string's character codes into a Stream<Character>.
2Collectors.groupingBy(c -> c, Collectors.counting()) groups equal characters together and counts how many are in each group.
3freq.entrySet().stream().max(Map.Entry.comparingByValue()) compares every entry by its count and keeps the single largest one, wrapped in an Optional.
4.orElseThrow() unwraps the result, since the string is known not to be empty here.

Grouping "success" produces counts for each character, and max() picks out 's': 3 as the largest entry.

💡

Key Point: Map.Entry.comparingByValue() is a ready-made comparator that compares entries purely by their value — no need to write a custom Comparator for this common case.

Complexity

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

Why: groupingBy() still visits every character once to build the map, and max() then makes one pass over its entries to find the largest.

Key Concepts

StreamCollectors.groupingBy()Map.Entry.comparingByValue()

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