Different Ways to Add Parentheses

Problem

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.

The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 10^4.

Example 1:

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

Example 2:

Input: expression = "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

Constraints:

1 <= expression.length <= 20
expression consists of digits and the operator '+', '-', and '*'.
All the integer values in the input expression are in the range [0, 99].

Solution

class Solution {
    public List<Integer> diffWaysToCompute(String expression) {
        return diffWaysToCompute(expression, new HashMap<>());
    }

    private List<Integer> diffWaysToCompute(String expression, Map<String, List<Integer>> map) {
        if (map.containsKey(expression)) {
            return map.get(expression);
        }
        List<Integer> values = new ArrayList<>();
        if (!hasOperator(expression)) {
            // base case
            values.add(Integer.parseInt(expression));
        } else {
            // Recursive case. DFS
            for (int i = 0; i < expression.length(); i++) {
                char symbol = expression.charAt(i);

                if (!Character.isDigit(symbol)) {
                    List<Integer> left = diffWaysToCompute(expression.substring(0, i), map);
                    List<Integer> right = diffWaysToCompute(expression.substring(i + 1), map);
                    for (Integer l : left) {
                        for (Integer r : right) {
                            switch (symbol) {
                                case '+':
                                    values.add(l + r);
                                    break;
                                case '-':
                                    values.add(l - r);
                                    break;
                                case '*':
                                    values.add(l * r);
                                    break;
                                default:
                                    break;
                            }
                        }
                    }
                }
            }
        }
        map.put(expression, values);
        return values;
    }

    private boolean hasOperator(String expression) {
        for (int i = 0; i < expression.length(); i++) {
            switch (expression.charAt(i)) {
                case '+':
                case '-':
                case '*':
                    return true;
                default:
                    break;
            }
        }
        return false;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).