## 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: $O(2^n)$
• Space Complexity: $O(1)$