# 241. Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1:

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

Example 2:

Input: "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

Approach 1: recursion.

# Code (Python)

Approach 1:

class Solution(object):
    def diffWaysToCompute(self, input):
        """
        :type input: str
        :rtype: List[int]
        """
        if input.isdigit():
            return [int(input)]
        result = []
        for i, char in enumerate(input):
            if char in '+-*':
                left = self.diffWaysToCompute(input[:i])
                right = self.diffWaysToCompute(input[i+1:])
                for l in left:
                    for r in right:
                        result.append(self.calc(l, r, char))
        return sorted(result)
    
    def calc(self, l, r, char):
        return {'+': lambda a,b: a+b, '-':lambda a,b: a-b, '*':lambda a,b: a*b}[char](l, r)

# Code (C++)

Approach 1:

class Solution {
private:
    unordered_map<string,vector<int>> tempVals;
    int doOp(int num1, int num2, char op) {
        switch (op)
        {
            case '+' : return num1 + num2;
            case '-' : return num1 - num2;
            case '*' : return num1 * num2;
        }
        return 0;
    }
public:
    vector<int> diffWaysToCompute(string input) {
        if (tempVals.find(input) != tempVals.end())
            return tempVals[input];
        vector<int> res;
        for (int i = 0; i < input.size(); ++i)
        {
            if (!isdigit(input[i]))
            {
                vector<int> left = diffWaysToCompute(input.substr(0, i));
                vector<int> right = diffWaysToCompute(input.substr(i+1, input.size()-i-1));
                for (int j = 0; j < left.size(); ++j)
                {
                    for (int k = 0; k < right.size(); ++k)
                    {
                        res.push_back(doOp(left[j], right[k], input[i]));
                    }
                }
            }
        }
        if (res.size() == 0)
            res.push_back(stoi(input));
        return res;
    }
};