# Check if characters of the given string can be rearranged to form a palindrome. # Counter is faster for long strings and non-Counter is faster for short strings.
// Time Complexity: O(n²) (due to substring creation and window reset) } else if (right == s.length() - 1 && !(map.containsKey(s.substring(left, right + 1)))) { map ...
Some results have been hidden because they may be inaccessible to you
Show inaccessible results