[LeetCode] 241. Different Ways to Add Parentheses

1. Question

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

2. Thoughts

divide the expression into left and right.
Using recursion, calculate left and right.

I think this could improve using memoization but couldn't do so.

3. Tips learned

4. My solution

 class Solution:
    def ComputeWith(self, left, right, operation):
        if operation == "+":
            return [x + y for x in left for y in right]
        elif operation == "-":
            return [x - y for x in left for y in right]
        elif operation == "*":
            return [x * y for x in left for y in right]

    def diffWaysToCompute(self, expression: str) -> List[int]:
        if expression.isdigit():
            return [int(expression)]
        oper = "-+*"
        result = []
        for i, char in enumerate(expression):
            if char in oper:
                left = self.diffWaysToCompute(expression[:i])
                right = self.diffWaysToCompute(expression[i + 1 :])
                result.extend(self.ComputeWith(left, right, char))

        return result

25/25 cases passed (34 ms)
Your runtime beats 84.68 % of python3 submissions
Your memory usage beats 86.71 % of python3 submissions (16.7 MB)

5. AI Solution and Improvements

Your solution effectively uses a divide-and-conquer strategy to solve the problem by breaking down the expression at each operator and recursively solving for each side. This approach is both intuitive and powerful for this type of problem. However, there are a couple of potential improvements and alternative strategies you might consider to optimize performance or approach the problem differently:

1. Memoization

One potential improvement to your current algorithm is to add memoization. This technique stores the results of expensive function calls and returns the cached result when the same inputs occur again. Since your function may be called multiple times with the same expression parameter (especially in cases with repeated sub-expressions), memoization can significantly reduce the number of computations.

2. Dynamic Programming

While the recursive approach with memoization is efficient, another way to think about this problem is using dynamic programming (DP). DP might be more complex to implement in this scenario due to the need to carefully manage the computation order and storage of intermediate results, but it can offer a bottom-up solution path that avoids the overhead of recursive calls.

3. Preprocessing to Separate Numbers and Operators

Your current solution parses the expression on the fly within the recursive calls. As an optimization, especially if the expression is long, you could preprocess the expression to separate numbers and operators into lists or arrays. This can make the handling of each part of the expression more efficient in the recursive calls, as you won't need to check each character to determine if it's a digit or an operator.

Implementation of Memoization

Here's an enhanced version of your solution with memoization added:

class Solution:
    def __init__(self):
        self.memo = {}

    def ComputeWith(self, left, right, operation):
        if operation == "+":
            return [x + y for x in left for y in right]
        elif operation == "-":
            return [x - y for x in left for y in right]
        elif operation == "*":
            return [x * y for x in left for y in right]

    def diffWaysToCompute(self, expression: str) -> [int]:
        if expression in self.memo:
            return self.memo[expression]
        
        if expression.isdigit():
            return [int(expression)]
        
        oper = "-+*"
        result = []
        for i, char in enumerate(expression):
            if char in oper:
                left = self.diffWaysToCompute(expression[:i])
                right = self.diffWaysToCompute(expression[i + 1:])
                result.extend(self.ComputeWith(left, right, char))
        
        self.memo[expression] = result
        return result

Final Thoughts

• Memoization is the easiest and most effective improvement for your existing solution.
• Dynamic Programming and Preprocessing could offer alternative approaches or further optimizations but may require more substantial changes to the algorithm structure.
• Always consider the trade-offs between code complexity and performance gains when choosing to implement these optimizations.