Different Ways to Add Parentheses

07/16/2016 Divide and Conquer

Question

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".

((2-1)-1) = 0
(2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: “23-45”

(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

Output: [-34, -14, -10, -10, 10]

Solution

Result: Accepted Time: 8 ms

Here should be some explanations.

class Solution {
public:
    vector<int> diffWaysToCompute(const string input) {
        vector<int> result;
        for(int i=0; i< input.size(); i++)
            if(!isdigit(input[i]))
                for(int a : diffWaysToCompute(input.substr(0, i)))
                    for(int b : diffWaysToCompute(input.substr(i+1)))
                        result.push_back(operate(a,b,input[i]));
        return result.size() ? result : vector<int>{stoi(input)};
    }
    inline int operate(int a,int b,char c)
    {
        if(c == '-') return a - b;
        if(c == '+') return a + b;
        return a*b;
    }
};

Complexity Analytics

  • Time Complexity:
  • Space Complexity: