Problem Statement   Correct parentheses sequences can be defined recursively as follows:
 The empty string "" is a correct sequence.
 If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.
 If "X" is a correct sequence, then "(X)" is a correct sequence.
 Each correct parentheses sequence can be derived using the above rules.
Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".
You are given a int[] p with n elements.
The elements of p are a permutation of the numbers 0 to n1.
Find strings s and t with the following properties:
 Both s and t are correct parentheses sequences.
 Each of them has exactly n characters.
 For each valid i, s[i] = t[ p[i] ].
Return one possible string s.
If there are multiple possibilities, you may return any of them.
If there is no solution, return "Impossible" instead.   Definition   Class:  ParenthesesAndPermutation  Method:  getSequence  Parameters:  int[]  Returns:  String  Method signature:  String getSequence(int[] p)  (be sure your method is public) 
    Constraints    p will contain between 2 and 50 elements, inclusive.    The length of p will be even.    p will be a permutation of 0 to (p1).   Examples  0)     Returns: "(())" 
We are looking for two correct parentheses sequences such that s[0]=t[2], s[1]=t[0], s[2]=t[3], and s[3]=t[1].
There are two parentheses sequences of length 4: "(())" and "()()". We can now argue as follows:
 Can s be "(())"? We can deduce that t must be "()()" which is a correct parentheses sequence, so this is a valid solution.
 Can s be "()()"? We can deduce that t must be "))((" which is not a correct parentheses sequence, so this is not a valid solution.
Therefore, the only valid solution is s = "(())".


 1)     Returns: "Impossible"  s and t must each be "()", but then s[0] != t[p[0]], so it is impossible to find such s and t. 

 2)     Returns: "(())(())"  Another valid solution is: s = t = "()()()()". 

 3)     4)    
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