In this article, we will discuss how to check if a string has all unique characters.

This is a frequently asked interview question. Here are few solutions in Java for this :

 

Solution 1 : Check if string has all unique characters using String library functions

package com.topjavatutorial;

public class UniqueCharactersDemo {

  public static void main(String[] args) {
    System.out.println(hasUniqueCharacters("hello"));
    
  }
  
  private static boolean hasUniqueCharacters(String str){
    for(char ch : str.toCharArray()){
      if(str.indexOf(ch) == str.lastIndexOf(ch))
        continue;
      else
        return false;
    }
    return true;
  }

}

Output :

false

 

Solution 2 : Check if string has unique characters using HashMap

  private static boolean checkUniqueUsingMap(String str){
    HashMap<Character,Integer> charMap = new HashMap<Character,Integer>();
    
    for(char ch : str.toCharArray()){
      if(charMap.containsKey(ch)){
        return false;
      }
      else
        charMap.put(ch, 1);
    }
    return true;
  }

 

Solution 3 : Compare each character to every other character without using an additional data structure

  public static boolean IsUniqueChars(String s) {
    for (int i = 0; i < s.length() - 1; i++) {
      for (int k = i + 1; k < s.length(); k++) {
        if (s.charAt(i) == s.charAt(k))
          return false;
      }
    }
    return true;
  }

This approach take O(n^2) time.
 

Solution 4 : Check if string has all unique characters without using an additional data structure in O(n) time

//Source: Cracking the Coding Interview Book
  public static boolean isUniqueChars(String str) {
    boolean[] char_set = new boolean[256];
    for (int i = 0; i < str.length(); i++) {
      // Get the ascii value of the character in str at position `i`.
      int val = str.charAt(i);
      // If char_set[val] has been set, that means that this character was
      // already present in the string.
      if (char_set[val])
        return false;
      // If the character hasn't been encountered yet, mark it as present
      // in the string
      char_set[val] = true;
    }
    // The string has unique chars if we reach this point
    return true;
  }

 

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