# Super Reduced String

See the original problem on HackerRank.

## Solutions

A solution changes the string by removing same adjacent elements.

Here is a C++ implementation:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19  #include #include #include using namespace std; int main() { string S; cin >> S; auto finder = [&]{ return adjacent_find(begin(S), end(S)); }; for (auto adjFind = finder(); adjFind != end(S); adjFind = finder()) { S.erase(adjFind, adjFind + 2); } cout << (S.empty() ? "Empty String" : S); } 

An alternative solution that does not modify the input uses an additional data structures:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27  #include #include #include #include using namespace std; int main() { string S; cin >> S; deque st; for (auto i=0; i

Haskell, grouping and counting adjacent equal characters, keeping only odd occurrences. The function is recursive and the base case is when original and reduced are equal.

  1 2 3 4 5 6 7 8 9 10  import Data.List (group) prettyPrint "" = "Empty String" prettyPrint s = s reduce s = if s' == s then s' else reduce s' where s' = map snd . (filter (odd . fst)) . map (\x -> (length x, head x)) $group s main = interact (prettyPrint . reduce)  Same, but with list comprehension:  1 2 3  reduce s = if s' == s then s' else reduce s' where s' = [v | (n, v) <- (\x -> (length x, head x)) <$> group s, odd n] 

Rust, using a stack.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15  fn reduce_string(s: String) -> String { s.chars().fold(String::new(), |mut acc, c| { match acc.pop() { None => acc.push(c), Some(l) => { if l != c { acc.push(l); acc.push(c); } } }; acc }) } 
We've worked on this challenge in these gyms: modena  padua  milan