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  1. Special String Again
  2. Discussions

Special String Again

  • oleksii_filonov
     + 4 comments

    Tried to simplify the code above and optimize a little bit by avoiding iterating over the repeatable symbols and jumping straight to the distinct by incrementing for loop counter and replacing sum with n*(n+1)/2 

    public static long substrCount(int length, String s) {
	long counter = 0;
	for (int i = 0; i < length; i++) {
		// if the current symbol is in the middle of palindrome, e.g. aba
		int offset = 1;
		while (i - offset >= 0 && i + offset < length && s.charAt(i - offset) == s.charAt(i - 1)
				&& s.charAt(i + offset) == s.charAt(i - 1)) {
			counter++;
			offset++;
		}
		// if this is repeatable characters aa
		int repeats = 0;
		while (i + 1 < length && s.charAt(i) == s.charAt(i + 1)) {
			repeats++;
			i++;
		}
		counter += repeats * (repeats + 1) / 2;
	}
	return counter + length;
}