Google codejam APAC Test practice round: Parentheses Order

Question

I spent one day solving this problem and couldn\'t find a solution to pass the large dataset.

Problem

An n parentheses sequence consists of n \"(\"s and n \"

Answer 1

Let S= any valid sequence of parentheses from n( and n) . Now any valid sequence S can be written as S=X+Y where

• X=valid prefix i.e. if traversing X from left to right , at any point of time, numberof'(' >= numberof')'
• Y=valid suffix i.e. if traversing Y from right to left, at any point of time, numberof'(' <= numberof')'

For any S many pairs of X and Y are possible.

For our example: ()(())

`()(())` =`empty_string + ()(())` 
         = `( + )(())`
         = `() + (())` 
         = `()( + ())` 
         = `()(( + ))` 
         = `()(() + )`
         = `()(()) + empty_string`

Note that when X=empty_string, then number of valid S from n( and n)= number of valid suffix Y from n( and n)

Now, Algorithm goes like this: We will start with X= empty_string and recursively grow X until X=S. At any point of time we have two options to grow X, either append '(' or append ')'

Let dp[a][b]= number of valid suffixes using a '(' and b ')' given X

nop=num_open_parenthesis_left ncp=num_closed_parenthesis_left

`calculate(nop,ncp)
{
  if dp[nop][ncp] is not known
  {
    i1=calculate(nop-1,ncp); // Case 1: X= X + "("
    i2=((nop=ncp, then after exhausting 1 ')' nop>ncp, therefore there can be no valid suffix*/
    dp[nop][ncp]=i1+i2;
  }
   return dp[nop][ncp];
}`

Lets take example,n=3 i.e. 3 ( and 3 ) Now at the very start X=empty_string, therefore

dp[3][3]= number of valid sequence S using 3( and 3 ) = number of valid suffixes Y from 3 ( and 3 )