Complexity

⏰ Time complexity: O(n)
🧺 Space complexity: O(1) (O(256) ⇨ O(1))

Method 3 - Sorting the Array and Then Finding Unique Chars

Here using the sorting algorithm to sort the string first, then compare every pair of neighboring characters to see if they are the same. The main time consuming are sorting array + neighborhood comparisions.

Code

Java

Java uses merge sort for object type arrays, and quicksort for primitive data types. Also for efficiency, it would switch to insertion sort when fewer than seven array elements are being sorted. So, time complexity in general is O(nlogn).

public boolean areAllCharsUnique(String s) {
 char[] chars = s.toCharArray();
 Arrays.sort(chars);
 for (int i = 0; i < chars.length - 1; i++) {
	 if

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