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 10<sup>4</sup>.

For example:

If the input is expression = "2-1-1" The expected output is [0,2] Because: ((2-1)-1) = 0 (2-(1-1)) = 2

If the input is expression = "2*3-4*5" The expected output is [-34,-14,-10,-10,10] Because: (2*(3-(45))) = -34 ((23)-(45)) = -14 ((2(3-4))5) = -10 (2((3-4)5)) = -10 (((23)-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.

Problem: Different Ways to Add Parentheses

Problem Description

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 10<sup>4</sup>.

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

Solution

This problem can be solved using recursion and dynamic programming (memoization) to avoid redundant calculations. The basic idea is to split the expression at each operator, recursively evaluate the left and right sub-expressions, and then combine the results based on the operator.

1. Naive Recursive Solution

def diffWaysToCompute_naive(expression):
    if expression.isdigit():
        return [int(expression)]

    results = []
    for i, char in enumerate(expression):
        if char in ['+', '-', '*']:
            left_results = diffWaysToCompute_naive(expression[:i])
            right_results = diffWaysToCompute_naive(expression[i+1:])

            for left in left_results:
                for right in right_results:
                    if char == '+':
                        results.append(left + right)
                    elif char == '-':
                        results.append(left - right)
                    else:
                        results.append(left * right)
    return results

2. Optimized Solution using Memoization

def diffWaysToCompute(expression):
    memo = {}
    
    def compute(expr):
        if expr in memo:
            return memo[expr]
        
        if expr.isdigit():
            return [int(expr)]
        
        results = []
        for i, char in enumerate(expr):
            if char in ['+', '-', '*']:
                left_results = compute(expr[:i])
                right_results = compute(expr[i+1:])

                for left in left_results:
                    for right in right_results:
                        if char == '+':
                            results.append(left + right)
                        elif char == '-':
                            results.append(left - right)
                        else:
                            results.append(left * right)
        memo[expr] = results
        return results

    return compute(expression)

Example Usage

expression1 = "2-1-1"
print(diffWaysToCompute(expression1))  # Output: [0, 2]

expression2 = "2*3-4*5"
print(diffWaysToCompute(expression2))  # Output: [-34, -14, -10, -10, 10]

Big(O) Runtime Analysis

• Naive Recursive Solution: The runtime complexity is exponential, roughly O(C(n)), where C(n) is the nth Catalan number. This is because for each operator, we split the expression and recursively compute the results. There are many overlapping subproblems, leading to redundant calculations.
• Memoized Solution: With memoization, we store the results of sub-expressions to avoid recomputation. The time complexity is significantly improved. If n is the length of the expression, we can say the runtime is O(n * number of sub-expressions), where the number of sub-expressions is still related to Catalan numbers but each sub-expression is computed only once.

Big(O) Space Usage Analysis

• Naive Recursive Solution: The space complexity is O(H), where H is the height of the recursion tree. In the worst case, it can be O(n).
• Memoized Solution: The space complexity is determined by the size of the memoization table, which stores results for different sub-expressions. It is also related to Catalan numbers but much better than the naive approach, since sub-expressions are only stored once. The space complexity can be considered O(number of sub-expressions).

Edge Cases

1. Empty Input: If the input string is empty, return an empty list.
2. Single Number: If the input string contains only a single number, return a list containing that number.
3. Invalid Characters: If the input string contains characters other than digits and '+', '-', '*', handle or raise an error as appropriate.
4. Large Numbers: If the numbers are too large and exceed integer limits, use appropriate data types or handle the overflow.
5. Division by Zero: If division is included as an operator, handle the case where the right operand is zero.