Скачать или смотреть 🔁 Find and Print Duplicate Characters in a String (Java)

Welcome to this walkthrough on how to find duplicate characters in a string using Java! This solution is simple, effective, and perfect for interviews, automation tools, or data validation tasks.

✅ What Is a Duplicate Character?

A character is considered a duplicate if it appears more than once in a string.

In this approach:

We treat uppercase and lowercase as the same (case-insensitive)

We ignore white spaces for cleaner results

🎯 Example

Input:
"Automation Testing"

After processing:
"a, t, o, i, n"

🧠 Step-by-Step Logic

Let’s break down how this program identifies duplicate characters:

🔹 Step 1: Convert the String to Lowercase

To make the comparison case-insensitive, the input string is converted entirely to lowercase. This ensures that characters like 'T' and 't' are treated the same.

🔹 Step 2: Create a Map to Track Character Frequency

We use a data structure that stores each character along with how many times it appears.

The key is the character

The value is the number of occurrences

🔹 Step 3: Iterate Through the Characters

We scan each character in the string:

If it’s already in the map, increase its count

If it’s not, add it to the map with a count of 1

🔹 Step 4: Identify and Print Duplicates

After building the frequency map, we go through each character and:

If its count is more than 1, we consider it a duplicate

These characters are printed as output

🔍 Example Walkthrough

Let’s analyze the string: "Automation Testing"

After converting to lowercase → "automation testing"

The frequency count becomes:

a → 2

u → 1

t → 3

o → 2

m → 1

i → 2

n → 2

e → 1

s → 1

g → 1

From this, the characters with more than one occurrence are:
a, t, o, i, n

✅ Output

The final printed results look like:

print removed duplicate characters: a

print removed duplicate characters: t

print removed duplicate characters: o

print removed duplicate characters: i

print removed duplicate characters: n

📌 Summary

Here’s what this program achieves:

Detects and prints characters that occur more than once

Uses a HashMap to store character counts

Ignores white spaces

Is case-insensitive for better accuracy

⏱️ Time & Space Complexity

Time Complexity: O(n), where n is the length of the input string

Space Complexity: O(k), where k is the number of unique characters

🔧 Optional Improvements

✅ Ignore spaces or punctuation by adding a condition

✅ Use a TreeMap instead of HashMap to get sorted output

✅ Return the duplicates in a list instead of printing

🔖 Hashtags

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