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Find Last Repeated Character in a String in Java

intermediate· Strings · String

Problem

A repeated character is one that appears two or more times in a string — the last one is whichever such character sits closest to the end.

Given a string, find the last character, reading right to left, that appears more than once.

Input

bookkeeper

Output

Last repeated character: e

Java Program

Java

import java.util.HashMap;
import java.util.Map;

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

        for (char c : str.toCharArray()) {
            freq.put(c, freq.getOrDefault(c, 0) + 1);
        }

        char result = '\0';
        for (int i = str.length() - 1; i >= 0; i--) {
            char c = str.charAt(i);
            if (freq.get(c) > 1) {
                result = c;
                break; // first repeated character found scanning from the end
            }
        }

        System.out.println("Last repeated character: " + result);
    }
}

▶ Run Code

Output

Last repeated character: e

Core Logic

Counting every character's total frequency first, then scanning the string backward, finds the last position holding a character that repeats.

How It Works

1A HashMap<Character, Integer> named freq tracks each character's total count across the whole string.
2freq.getOrDefault(c, 0) reads the current count, or 0 if unseen, then freq.put(c, ...) stores the incremented count.
3A second loop walks the string backward, from index str.length() - 1 down to 0.
4The first character encountered in that backward walk whose freq.get(c) is greater than 1 is the answer, printed immediately with break.

In "bookkeeper", the last character 'r' has a count of just 1, so the scan continues backward until it reaches 'e', which has a count of 3.

💡

Key Point: This is the mirror image of finding the first repeated character — same frequency map, but the second scan runs backward instead of forward.

Complexity

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

Why: The first pass builds a frequency map holding one entry per distinct character, and the second pass re-scans the string backward looking up each character's count.

Key Concepts

HashMapgetOrDefault()reverse scan

Approach 2: Java 8

Java

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

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

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

        // Walks the string's indices in reverse, stopping at the first repeated character
        char result = IntStream.range(0, str.length())
            .map(i -> str.length() - 1 - i)
            .mapToObj(str::charAt)
            .filter(c -> freq.get(c) > 1)
            .findFirst()
            .orElseThrow();

        System.out.println("Last repeated character: " + result);
    }
}

▶ Run Code

Output

Last repeated character: e

Core Logic

The same mirrored-index trick used to reverse an array or string can walk the string's indices backward as a stream, stopping at the first repeated character it finds.

How It Works

1str.chars().mapToObj(c -> (char) c) and Collectors.groupingBy() build the frequency map exactly like the manual version.
2IntStream.range(0, str.length()).map(i -> str.length() - 1 - i) generates the string's indices in reverse order.
3.mapToObj(str::charAt) converts each reversed index into its character.
4.filter(c -> freq.get(c) > 1).findFirst() keeps only repeated characters and takes the first one found in this backward order.

Walking "bookkeeper"'s indices in reverse visits 'r' first (count 1, skipped), then 'e' (count 3, matched).

💡

Key Point: IntStream.range(...).map(i -> length - 1 - i) is the same reversed-index pattern used elsewhere on this site to walk a sequence backward without building a separate reversed copy first.

Complexity

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

Why: groupingBy() still visits every character once to build the map, and the reversed-index stream then makes one more pass looking for the first repeated character.

Key Concepts

StreamIntStreamfilter()findFirst()

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