Given a string S of length N, consisting of only opening ‘(‘ and closing ‘)‘ parenthesis. The task is to find all indices ‘K‘ such that S[K…N-1] + S[0…K-1] is a regular parenthesis.
A regular parentheses string is either empty (“”), “(” + str1 + “)”, or str1 + str2, where str1 and str2 are regular parentheses strings.
For example: “”, “()”, “(())()”, and “(()(()))” are regular parentheses strings.
Input: str = “)()(”
For K = 1, S = ()(), which is regular.
For K = 3, S = ()(), which is regular.
Input: S = “())(”
For K = 3, S = (()), which is regular.
Naive Approach: The naive approach is to split the given string str at every possible index(say K) and check whether str[K, N-1] + str[0, K-1] is palindromic or not. If yes then print that particular value of K.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The idea is to observe that if at any index(say K) where the count of closing brackets is greater than the count of opening brackets then that index is the possible index of splitting the string. Below are the steps:
- The partition is only possible when the count the number of opening brackets must be equal to the number of closing brackets. Else we can’t form any partition to balanced the parenthesis.
- Create an auxiliary array(say aux) of size length of the string.
- Traverse the given string if character at any index(say i) is ‘(‘ then update aux[i] to 1 else update strong>aux[i] to -1.
- The frequency of the minimum element in the above auxiliary array is the required number of splitting(say at index K) to make S[K…N-1] + S[0…K-1] a regular parenthesis string.
Below is the implementation of the above approach:
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(N), where N is the length of the string.
- Find a valid parenthesis sequence of length K from a given valid parenthesis sequence
- Count of cyclic permutations having XOR with other binary string as 0
- Count number of ways to convert string S to T by performing K cyclic shifts
- Number of balanced parenthesis substrings
- InfyTQ 2019 : Find the position from where the parenthesis is not balanced
- Reverse substrings between each pair of parenthesis
- Identify and mark unmatched parenthesis in an expression
- Calculate weight of parenthesis based on the given conditions
- Find maximum depth of nested parenthesis in a string
- Minimum moves to reach from i to j in a cyclic string
- Check if a decreasing Array can be sorted using Triple cyclic shift
- Sort a string lexicographically using triple cyclic shifts
- Find initial sequence that produces a given Array by cyclic increments upto index P
- Count number of indices such that s[i] = s[i+1] : Range queries
- Count maximum occurrence of subsequence in string such that indices in subsequence is in A.P.
- Count of maximum occurring subsequence using only those characters whose indices are in GP
- Find the starting indices of the substrings in string (S) which is made by concatenating all words from a list(L)
- Find indices of all occurrence of one string in other
- Sum of all the Composite Numbers from Odd indices of the given array
- Count distinct regular bracket sequences which are not N periodic
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