regex match longest substring with equal first and last char - regex

Question

/(\w)(\w*)\1/
For this string:"mgntdygtxrvxjnwksqhxuxtrv" I match "txrvxjnwksqhxuxt" (using Ruby), but not the even longer valid substring "tdygtxrvxjnwksqhxuxt".

Answer 1

For a given string, here are two ways to find the longest substring that begins and ends with the same character.
Suppose
str = "mgntdygtxrvxjnwksqhxuxtrv"
Use a regular expression
r = /(.)(?=(.*\1))/
str.gsub(r).map { $1 + $2 }.max_by(&:length)
#=> "tdygtxrvxjnwksqhxuxt".
When, as here, the regular expression contains capture groups, it may be more convenient to use String#gsub without a second argument or block (in which case it returns an enumerator, which can be chained) than String#scan (" If the pattern contains groups, each individual result is itself an array containing one entry per group.") Here gsub performs no substitutions; it merely generates matches of the regular expression.
The regular expression can be made self-documenting by writing it in free-spacing mode.
r = /
(.) # match any char and save to capture group 1
(?= # begin a positive lookahead
(.*\1) # match >= 0 characters followed by the contents of capture group 1
) # end the postive lookahead
/x # free-spacing regex definition mode
The following intermediate calculation is performed:
str.gsub(r).map { $1 + $2 }
#=> ["gntdyg", "ntdygtxrvxjn", "tdygtxrvxjnwksqhxuxt", "txrvxjnwksqhxuxt",
# "xrvxjnwksqhxux", "rvxjnwksqhxuxtr", "vxjnwksqhxuxtrv", "xjnwksqhxux",
# "xux"]
Notice that this does not enumerate all substrings beginning and ending with the same character (because .* is greedy). It does not generate, for example, the substring "xrvx".
Do not use a regular expression
v = str.each_char.with_index.with_object({}) do |(c,i),h|
if h.key?(c)
h[c][:size] = i - h[c][:start] + 1
else
h[c] = { start: i, size: 1 }
end
end.max_by { |_,h| h[:size] }.last
str[v[:start], v[:size]]
#=> "tdygtxrvxjnwksqhxuxt"