241 Different Ways To Add Parentheses

Different Ways to Add Parentheses

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.

The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 104.

 

Example 1:

**Input:** expression = "2-1-1"
**Output:** [0,2]
**Explanation:**
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

**Input:** expression = "2*3-4*5"
**Output:** [-34,-14,-10,-10,10]
**Explanation:**
(2*(3-(4*5))) = -34 
((2*3)-(4*5)) = -14 
((2*(3-4))*5) = -10 
(2*((3-4)*5)) = -10 
(((2*3)-4)*5) = 10

 

Constraints:

1 <= expression.length <= 20
expression consists of digits and the operator '+', '-', and '*'.
All the integer values in the input expression are in the range [0, 99].
The integer values in the input expression do not have a leading '-' or '+' denoting the sign.

from typing import List

class Solution:
    def __init__(self):
        self.cache = {}
    
    def diffWaysToCompute(self, expression: str) -> List[int]:
        # If the result is already cached, return it
        if expression in self.cache:
            return self.cache[expression]
        
        # This will store all the possible results
        res = []
        
        # Traverse the expression and evaluate based on operators
        for i, c in enumerate(expression):
            if c in "+-*":
                # Divide the problem into left and right parts
                left = self.diffWaysToCompute(expression[:i])
                right = self.diffWaysToCompute(expression[i+1:])
                
                # Calculate all combinations of left and right parts with the current operator
                for l in left:
                    for r in right:
                        if c == "+":
                            res.append(l + r)
                        elif c == "-":
                            res.append(l - r)
                        elif c == "*":
                            res.append(l * r)
        
        # Base case: if there are no operators, return the number itself
        if not res:
            res = [int(expression)]
        
        # Cache the result
        self.cache[expression] = res
        return res