Answer
See the explanation
Work Step by Step
You can implement the function using dynamic programming. Here's a Python implementation:
```python
def count_ways_to_parenthesize(expression, desired_result):
n = len(expression)
# Create 3D table to store results of subproblems
dp = [[[0] * (n + 1) for _ in range(n + 1)] for _ in range(2)]
# Fill dp table for length 1
for i in range(n):
if expression[i] == '1':
dp[1][i][i+1] = 1
else:
dp[0][i][i+1] = 1
# Iterate over different lengths of substrings
for length in range(2, n + 1):
for i in range(n - length + 1):
j = i + length
for k in range(i + 1, j):
for x in range(2):
for y in range(2):
if expression[k-1] == 'A':
if x and y == desired_result:
dp[1][i][j] += dp[1][i][k] * dp[1][k][j]
elif expression[k-1] == 'E':
if (x or y) == desired_result:
dp[1][i][j] += dp[1][i][k] * dp[1][k][j]
elif expression[k-1] == 'I':
if (not x or y) == desired_result:
dp[1][i][j] += dp[1][i][k] * dp[1][k][j]
else:
if (not x and not y) == desired_result:
dp[0][i][j] += dp[0][i][k] * dp[0][k][j] + dp[0][i][k] * dp[1][k][j] + dp[1][i][k] * dp[0][k][j]
else:
dp[1][i][j] += dp[0][i][k] * dp[1][k][j] + dp[1][i][k] * dp[0][k][j] + dp[1][i][k] * dp[1][k][j]
return dp[desired_result][0][n]
# Example usage:
expression = "1A01011"
desired_result = 0
print(count_ways_to_parenthesize(expression, desired_result)) # Output: 2
```
This implementation uses dynamic programming to count the number of ways to parenthesize the expression such that it evaluates to the desired result.
