PROBLEM LINK:
Author: Ankit Srivastava and Uttam Kanodia
Tester: Roman Rubanenko
Editorialist: Amit Pandey
DIFFICULTY:
Easy-Medium
PREREQUISITES:
Basic programming
PROBLEM:
Given a string containing '(' and ')' only. Find the longest such subsequence of the given string that is non-regular. Amongst all such distinct answers, output the lexicographically $K^{th}$ amongst them.
QUICK EXPLANATION:
If the given string in not balanced, then the string itself is the largest non-regular(unbalanced) pattern. On the contrary, if the given string is regular(balanced), then all subsequnces having length $(N-1)$ will be non-regular and there is a pattern in their lexicographic ordering.
EXPLANATION:
The problem can be broadly classified into two cases:
Case I: Given parenthesis string is not balanced, which can be checked using stack data structure described here. In this case, the largest unbalanced sub-sequence will be then the given string itself. So if $K=1$, then the answer will be the given string, $-1$ otherwise.
Case II: Given parenthesis string is the balanced one. Consider that $L$ is the length of the given string, then there will $L$ sub-sequences of length $(L-1)$, and each of them will be unbalanced. This is because, in any sub-sequence, number of opening and closing parenthesis will not be the same.
How to find the number of distinct unbalanced sub sequences?
Consider a string $S$ = "(())", what will be number of distinct unbalanced sub-sequences? it can be easily claimed that if we delete $S[0]$ or $S[1]$, we will get same string i.e. "())". So, it can be seen easily that if we delete any character from a contiguous chunk of the characters, we will get same unbalanced sub-sequence. It can be formally written as follows:
Consider the any contiguous chunk of same characters in the given string, Suppose $S[i] = S[i+1] = S[i+2] =\ ....\ = S[j] =\ '('$ or $')'$. If we delete any character between $i^{th}$ and $j^{th}$ positions (both inclusive), it will produce the same sub-sequence, because deleting any of them will replace $(j-i)+1$ number of same characters by $(j-i)$ number of same characters. So number of distinct sub-sequences of length $(L-1)$ will be number of different contiguous chunks of same characters. For example, string $(())((()))$ will have $4$ distinct sub-sequence of length $9$, and each of them will be unbalanced.
How to do lexicographic ordering of those sub-sequences?
Suppose that the string which we generate by deleting a character from the $i^{th}$ chunk (0-indexed) is $L_i$. As our given string is balanced, it will consist of one or more opening parentheses and one or more closing parentheses alternatively (starting with opening). So, to produce $L_{2n}$ we need to delete an opening parenthesis, and to produce $L_{2n+1}$, we need to delete a closing parenthesis. Now, let me claim that the lexicographic ordering:
$$L_{2i+1} < L_{2j+1}, \hspace{2 mm} L_{2i} > L_{2j} \hspace{2 mm} for \hspace{2 mm} (i < j) $$ $$L_{2i+1} < L_{2j} \hspace{2 mm} \text{for all} \hspace{2 mm}(i,j) $$
So for string $S = ''(())()((()))(())"$, the lexicographic order will be $L_1 < L_3 < L_5 < L_7 < L_6 < L_4 < L_2 < L_0$. To produce lexicographically $3^{rd}$ string we need to produce $L_5$ by deleting any of the bold characters : "(())()((()))(())".
Proof: Consider the proof $L_{2i+1} < L_{2j+1},\hspace{2 mm} \forall \hspace{2 mm} (i < j)$. Others can be proved in the same manner.
Consider that the $i^{th}$ chunk has $A_i$ number of characters. Now consider the $(A_1 + A_2 + .... A_{2i+1})^{th}$ character in $L_{2i+1}$ and $L_{2j+1}$, it will be ')' in $L_{2j+1}$, as $i < j$, and it will remain the same as in the original string, but it will be '(' in $L_{2i+1}$ and all previous characters will be the same in both of the strings. As opening parenthesis is lexicographically smaller than closing parenthesis, hence $L_{2i+1} < L_{2j+1}$.
Solution:
Setter's solution can be found here
Tester's solution can be found here