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241. Different Ways to Add Parentheses

Description

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 104.

 

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].
  • The integer values in the input expression do not have a leading '-' or '+' denoting the sign.

Solutions

Solution 1

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class Solution:
    def diffWaysToCompute(self, expression: str) -> List[int]:
        @cache
        def dfs(exp):
            if exp.isdigit():
                return [int(exp)]
            ans = []
            for i, c in enumerate(exp):
                if c in '-+*':
                    left, right = dfs(exp[:i]), dfs(exp[i + 1 :])
                    for a in left:
                        for b in right:
                            if c == '-':
                                ans.append(a - b)
                            elif c == '+':
                                ans.append(a + b)
                            else:
                                ans.append(a * b)
            return ans

        return dfs(expression)

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class Solution {
    private static Map<String, List<Integer>> memo = new HashMap<>();

    public List<Integer> diffWaysToCompute(String expression) {
        return dfs(expression);
    }

    private List<Integer> dfs(String exp) {
        if (memo.containsKey(exp)) {
            return memo.get(exp);
        }
        List<Integer> ans = new ArrayList<>();
        if (exp.length() < 3) {
            ans.add(Integer.parseInt(exp));
            return ans;
        }
        for (int i = 0; i < exp.length(); ++i) {
            char c = exp.charAt(i);
            if (c == '-' || c == '+' || c == '*') {
                List<Integer> left = dfs(exp.substring(0, i));
                List<Integer> right = dfs(exp.substring(i + 1));
                for (int a : left) {
                    for (int b : right) {
                        if (c == '-') {
                            ans.add(a - b);
                        } else if (c == '+') {
                            ans.add(a + b);
                        } else {
                            ans.add(a * b);
                        }
                    }
                }
            }
        }
        memo.put(exp, ans);
        return ans;
    }
}

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class Solution {
public:
    vector<int> diffWaysToCompute(string expression) {
        return dfs(expression);
    }

    vector<int> dfs(string exp) {
        if (memo.count(exp)) return memo[exp];
        if (exp.size() < 3) return {stoi(exp)};
        vector<int> ans;
        int n = exp.size();
        for (int i = 0; i < n; ++i) {
            char c = exp[i];
            if (c == '-' || c