Calculate the number of ways to parenthesize a boolean expression in Python

Question:

Calculate the number of ways to parenthesize a boolean expression in Python

Solution:

def countWays(self, n, s):

# code here

dp=[[[0,0] for _ in range(n)]for _ in range(n)]

for i in range(0,n,2):

if s[i]=='T':

dp[i][i][1]=1

else:

dp[i][i][0]=1

for x in range(3,n+1,2):

for i in range(0,n-x+1,2):

j=i+x-1

t=(x-1)//2

for k in range(1,t+1):

k2=2*k

if s[i+k2-1]=='&':

dp[i][j][1]+=dp[i][i+k2-2][1]*dp[i+k2][j][1]

dp[i][j][0]+=dp[i][i+k2-2][1]*dp[i+k2][j][0]+dp[i][i+k2-2][0]*dp[i+k2][j][1]+dp[i][i+k2-2][0]*dp[i+k2][j][0]

elif s[i+k2-1]=='|':

dp[i][j][0]+=dp[i][i+k2-2][0]*dp[i+k2][j][0]

dp[i][j][1]+=dp[i][i+k2-2][0]*dp[i+k2][j][1]+dp[i][i+k2-2][1]*dp[i+k2][j][0]+dp[i][i+k2-2][1]*dp[i+k2][j][1]

else:

dp[i][j][0]+=dp[i][i+k2-2][0]*dp[i+k2][j][0]+dp[i][i+k2-2][1]*dp[i+k2][j][1]

dp[i][j][1]+=dp[i][i+k2-2][1]*dp[i+k2][j][0]+dp[i][i+k2-2][0]*dp[i+k2][j][1]

return dp[0][n-1][1]%1003

Explenation:

This Python code defines a function countWays within a class Solution. The function aims to count the number of ways to parenthesize a boolean expression to achieve the desired outcome. Here's a brief explanation:

The function takes two parameters: n, which represents the length of the boolean expression, and s, which is the boolean expression as a string.
It initializes a 3D dynamic programming array dp to store the results of subproblems.
The first loop initializes base cases for single-character expressions (odd-length substrings). It assigns 1 to dp[i][i][1] if s[i] is 'T' (true), and to dp[i][i][0] if s[i] is 'F' (false).
The second loop iterates over all possible subexpressions of length greater than 1 (even-length substrings). Within this loop:
It calculates the number of ways to parenthesize the current subexpression based on the operators '&' (and), '|' (or), and '^' (xor).
It updates the values in the dp array accordingly based on the number of ways to parenthesize the subexpressions on both sides of the current operator.

Finally, it returns the result modulo 1003.

Overall, the code efficiently calculates the number of ways to parenthesize the given boolean expression using dynamic programming techniques.

Answered by: >21r01a0528

Credit: >GeekforGeeks

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