BRCKTS - Brackets
We will call a bracket word any word constructed out of two sorts of characters: the opening bracket "(" and the closing bracket ")". Among these words we will distinguish correct bracket expressions. These are such bracket words in which the brackets can be matched into pairs such that
- every pair consists of an opening bracket and a closing bracket appearing further in the bracket word
- for every pair the part of the word between the brackets of this pair has equal number of opening and closing brackets
- replacement -- changes the i-th bracket into the opposite one
- check -- if the word is a correct bracket expression
Write a program which
- reads (from standard input) the bracket word and the sequence of operations performed,
- for every check operation determines if the current bracket word is a correct bracket expression,
- writes out the outcome (to standard output).
Ten test cases (given one under another, you have to process all!). Each of the test cases is a series of lines. The first line of a test consists of a single number n (1<=n<=30000) denoting the length of the bracket word. The second line consists of n brackets, not separated by any spaces. The third line consists of a single number m -- the number of operations. Each of the following m lines carries a number k denoting the operation performed. k=0 denotes the check operation, k>0 denotes replacement of k-th bracket by the opposite.
For every test case your program should print a line:
where i is replaced by the number of the test and in the following lines, for every check operation in the i-th test your program should print a line with the word YES, if the current bracket word is a correct bracket expression, and a line with a word NO otherwise. (There should be as many lines as check operations in the test.)
Input: 4 ()(( 4 4 0 2 0 [and 9 test cases more] Output: Test 1: YES NO [and 9 test cases more]Warning: large Input/Output data, be careful with certain languages
@abid1729 No. See https://en.wikipedia.org/wiki/Context-free_grammar#Well-formed_parentheses
This was straightforward problem if you can figure out how to merge left and right halves of the string and figure out if its valid or not.
I have tried each and everything , still my code is giving TLE.
@lokesh_2052 I think you should find the min sum in the segment tree. For example ))(( is -1 -1 1 1, and it's sum-segment tree is based on -1 -2 -1 0, then find min in the segment tree.
I did this in 2.42s. I could not do myself . I find confusion ))(( is right or wrong . this is wrong!!!!!!
Just a simple Segment Tree construction....
Solved using segment tree. Nice problem!
can this question be solved using stack???
@abid1729 no its not but when you change the characters at 3 and 4 it becomes correct ()()()
Use "puts" or "printf".