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".

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

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(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]

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result = visit(input);
	    return result;
    }
private:
    int findSymbol(string input, int pos)
    {
	    int a = input.find('+', pos);
	    if (a == -1)
	    {
		    a = INT_MAX;
	    }
	    int b = input.find('-', pos);
	    if (b == -1)
	    {
		    b = INT_MAX;
	    }
	    int c = input.find('*', pos);
	    if (c == -1)
	    {
		    c = INT_MAX;
	    }

	    int result = min(a, min(b, c));
	    return result;
    }
    
    vector<int> visit(string input)
    {
	    int n = input.length();
    	vector<int> result;
	    if (findSymbol(input, 0) == INT_MAX)
	    {
		    result.push_back(atoi(input.c_str()));
	    }
	    else
	    {
		    int pos = 0;
		    int index = 0;
		    while ((index = findSymbol(input, pos)) != INT_MAX)
		    {
			    vector<int> a = visit(input.substr(0, index));
			    vector<int> b = visit(input.substr(index+1));
			    for (int i = 0; i < a.size(); i++)
			    {
				    for (int j = 0; j < b.size(); j++)
				    {
					    if (input[index] == '+')
					    {
						    result.push_back(a[i] + b[j]);
					    }
					    else if (input[index] == '-')
					    {
						    result.push_back(a[i] - b[j]);
					    }
					    else
					    {
						    result.push_back(a[i] * b[j]);
					    }
				    }
			    }
			    pos = index + 1;
		    }
	    }

	    return result;
    }
};