Java find the index of string based on the utf-8 encoding index

String text = "un’accogliente villa del."; // Unicode text
text = Normalizer.normalize(text, Form.NFC); // Normalize text

byte[] bytes = text.getBytes(StandardCharsets.UTF_8); // Index 5 UTF-8; 1 byte
char[] chars = text.toCharArray();                    // Index 3 UTF-16; 2 bytes (indexOf)
int[] codePoints = text.codePoints().toArray();       // Index 3 UTF-32; 4 bytes

int charIndex = text.indexOf("accogliente");
int codePointIndex = (int) text.substring(0, charIndex).codePoints().count();
int byteIndex = text.substring(0, charIndex).getBytes(StandardCharsets.UTF_8).length;

UTF-32 is the Unicode code points, the numbering of all symbols with U+XXXX where there maybe more (or less) than 4 hexadecimal digits.

Text normalisation is needed as é could be one code point, or two code points, a zero-width ´ followed by a e.

The question of UTF-8 byte index to UTF-16 char index:

int charIndex = new String(text.getBytes(StandardCharsets.UTF_8),
                           0, byteIndex, StandardCharsets.UTF_8).length();

Below code will return output as 3 am i missing something in your question?

String text="un’accogliente villa del.";
text.indexOf("accogliente");

Well assuming that this startIndex can only be a letter (ASCII one), you could do:

String text = "un’accogliente villa del.";
char c = text.charAt(5);
String normalized = Normalizer.normalize(text, Normalizer.Form.NFD);
normalized = normalized.replaceAll("[^\\p{ASCII}]", " ");

Pattern p = Pattern.compile("\\p{L}*?" + c + "\\p{L}*?[$|\\s]");
Matcher m = p.matcher(normalized);

if (m.find()) {
     System.out.println(m.start(0));
}