Determine Unique Characters in String

Written by:Fatih KABAKCI

The question is to determine if a string has all unique characters. First solution easily comes up to mind is to implement brute force solution of this problem. Get each character in an upper loop, then compare rest of other characters in an inner loop. Let's look at the the first solution which its complexity is O(n2) in below.

Brute Force

/**
 * 
 * @author fkabakci
 * Find out if a string has all unique characters -- Brute Force O(n2) solution
 */
class Solution {
	public static boolean isUnique(String str) {
		for(int i = 0; i < str.length(); i++) {
			for(int j = i + 1; j < str.length(); j++) {
				if(str.charAt(i) == str.charAt(j))
					return false;
			}
		}
		return true;
	}

	public static void main(String[] args) {
		String computer = "Computer";
		String printer = "Printer";

		System.out.println(isUnique(computer));
		System.out.println(isUnique(printer));
	}
}

O(n) Solution

We know that all printable ASCII characters are up to 127. So all possible characters are ordered till ASCII 127. Therefore, while we get each character, we could have a boolean buffer array that stores those characters whether they appeared or not as true and false values. The indices of this buffer array uses the ASCII code of the characters in the string.

/**
 * 
 * @author fkabakci
 * Find out if a string has all unique characters -- O(n) solution
 */
class Solution {
	public static boolean isUnique(String str) {
		int length = str.length();
		// ASCII
		if (length > 128) {
			return false;
		}
		boolean[] flag = new boolean[128];
		for (int i = 0; i < length; i++) {
			int code = str.codePointAt(i);
			if (flag[code])
				return false;
			flag[code] = true;
		}
		return true;
	}

	public static void main(String[] args) {
		String computer = "Computer";
		String printer = "Printer";

		System.out.println(isUnique(computer));
		System.out.println(isUnique(printer));
	}
}

Now, we have reached out the O(n) solution by just iterating the array once.