Regular Expression to match 3 or more Consecutive Sequential Characters and Consecutive Identical Characters

Question

I need regular expressions to match the below cases.

1. 3 or more consecutive sequential characters/numbers; e.g. 123, abc, 789, pqr, etc.
2. 3 or more cons

Answer 1

I don't think you can use regex for the first case. The second case is easy though:

Pattern pattern = Pattern.compile("([a-z\\d])\\1\\1", Pattern.CASE_INSENSITIVE);

Since \\1 represents part matched by group 1 this will match any sequence of three identical characters that are either within the range a-z or are digits (\d).

Answer 2

for the second question:

\\b([a-zA-Z0-9])\\1\\1+\\b

explanation:

\\b               : zero-length word boundary
  (               : start capture group 1
    [a-zA-Z0-9]   : a letter or a digit
  )               : end group
  \\1             : same character as group 1
  \\1+            : same character as group 1 one or more times
\\b               : zero-length word boundary

Answer 3

All put together:

([a-zA-Z0-9])\1\1+|(abc|bcd|cde|def|efg|fgh|ghi|hij|ijk|jkl|klm|lmn|mno|nop|opq|pqr|qrs|rst|stu|tuv|uvw|vwx|wxy|xyz|012|123|234|345|456|567|678|789)+

3 or more consecutive sequential characters/numbers; e.g. 123, abc, 789, pqr, etc.

(abc|bcd|cde|def|efg|fgh|ghi|hij|ijk|jkl|klm|lmn|mno|nop|opq|pqr|qrs|rst|stu|tuv|uvw|vwx|wxy|xyz|012|123|234|345|456|567|678|789)+

3 or more consecutive identical characters/numbers; e.g. 111, aaa, bbb, 222, etc.

([a-zA-Z0-9])\1\1+

https://regexr.com/4727n

This also works:

(?:(?:0(?=1)|1(?=2)|2(?=3)|3(?=4)|4(?=5)|5(?=6)|6(?=7)|7(?=8)|8(?=9)){2,}\d|(?:a(?=b)|b(?=c)|c(?=d)|d(?=e)|e(?=f)|f(?=g)|g(?=h)|h(?=i)|i(?=j)|j(?=k)|k(?=l)|l(?=m)|m(?=n)|n(?=o)|o(?=p)|p(?=q)|q(?=r)|r(?=s)|s(?=t)|t(?=u)|u(?=v)|v(?=w)|w(?=x)|x(?=y)|y(?=z)){2,}[[:alpha:]])|([a-zA-Z0-9])\1\1+

https://regex101.com/r/6fXC9u/1