# Different Ways to Add Parentheses in C++

Suppose we have a string of numbers and operators, we have to find all possible results from computing all the different possible ways to group the numbers and operators. Here the valid operators are +, - and *. So if the input is like “2*3-4*5”, then the output will be [-34, -14, -10, -10, 10]. This is because −

• (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

To solve this, we will follow these steps −

• Define a map called a memo.

• Define a method called solve(). This will take the input string as input.

• create an array called ret

• if the memo has input, then return memo[input]

• for i in range 0 to the size of input string −

• if input[i] is any supported operator, then

• an array part1 := solve(substring of input from 0 to i - 1)

• an array part2 := solve(substring of input from i to end of string)

• for j in range 0 to size of part1

• for k in range 0 to size of part2

• if input[i] is addition, then

• perform part[j] + part[k] and add into ret

• if input[i] is multiplication, then

• perform part[j] * part[k] and add into ret

• if input[i] is subtraction, then

• perform part[j] - part[k] and add into ret

• if ret is empty, then return input string as an integer

• memo[input] := ret, and return ret

## Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
map <string, vector<int>> memo;
vector<int> diffWaysToCompute(string input) {
vector <int> ret;
if(memo.count(input)) return memo[input];
for(int i = 0; i < input.size(); i++){
if(input[i] == '+' || input[i] == '*' || input[i] == '-'){
vector <int> part1 = diffWaysToCompute(input.substr(0, i));
vector <int> part2 = diffWaysToCompute(input.substr(i + 1));
for(int j = 0; j < part1.size(); j++ ){
for(int k = 0; k < part2.size(); k++){
if(input[i] == '+'){
ret.push_back(part1[j] + part2[k]);
}
else if(input[i] == '*'){
ret.push_back(part1[j] * part2[k]);
} else {
ret.push_back(part1[j] - part2[k]);
}
}
}
}
}
if(ret.empty()){
ret.push_back(stoi(input));
}
return memo[input] = ret;
}
};
main(){
Solution ob;
print_vector(ob.diffWaysToCompute("2*3-4*5"));
}

## Input

"2*3-4*5"

## Output

[-34, -10, -14, -10, 10, ]