Question 5 : How to find duplicate characters in String in java?

 Solution:  Here is a solution to find duplicate characters in String.

1. Create a HashMap and the character of String will be inserted as key and its count as value.
2. If HashMap already contains char, increase its count by 1, else put char in HashMap.
3. If the value of Char is more than 1, that means it is a duplicate character in that String.

Java Program to find duplicate Characters in a String:

import java.util.HashMap;  
import java.util.Map;  
import java.util.Set;  
   
public class DuplicateCharFinder {  
    public void findIt(String str) {  
        Map<Character, Integer> baseMap = new HashMap<Character, Integer>();  
        char[] charArray = str.toCharArray();  
        for (Character ch : charArray) {  
            if (baseMap.containsKey(ch)) {  
                baseMap.put(ch, baseMap.get(ch) + 1);  
            } else {  
                baseMap.put(ch, 1);  
            }  
        }  
        Set<Character> keys = baseMap.keySet();  
        for (Character ch : keys) {  
            if (baseMap.get(ch) > 1) {  
                System.out.println(ch + "  is " + baseMap.get(ch) + " times");  
            }  
        }  
    }  
   
    public static void main(String a[]) {  
        DuplicateCharFinder dcf = new DuplicateCharFinder();  
        dcf.findIt("India is my country");  
    }  
}

Python code 

def find_duplicates(s: str):
    freq = {}

    for ch in s:
        freq[ch] = freq.get(ch, 0) + 1

    for ch, count in freq.items():
        if count > 1:
            print(f"{ch} -> {count}")


if __name__ == "__main__":
    find_duplicates("cloudtechtwitter")

📌 Final Comparison

  
    
    
    
    
  

  

    
Approach
    
Time Complexity
    
Space Complexity
    
Best Use
  

  

    

      Frequency Map (HashMap / Dictionary)
    
    

      O(n)
    
    

      O(k)
    
    

      ⭐ Best & general solution
    
  

  

    

      Sorting characters
    
    

      O(n log n)
    
    

      O(n)
    
    

      Simple but slower
    
  

  

    

      indexOf & lastIndexOf
    
    

      O(n²)
    
    

      O(1)
    
    

      ❌ Avoid for large input
    
  

  

    

      ASCII frequency array
    
    

      O(n)
    
    

      O(1)
    
    

      Fastest for ASCII only