Different Ways to Add Parentheses LeetCode Solution

Different Ways to Add Parentheses 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.

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].

Different Ways to Add Parentheses Solutions

✅Time: O(2n→n)
✅Space: O(n)

C++

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

 private:
  vector<int> ways(const string& s, unordered_map<string, vector<int>>&& memo) {
    if (memo.count(s))
      return memo[s];

    vector<int> ans;

    for (int i = 0; i < s.length(); ++i)
      if (ispunct(s[i]))
        for (const int a : ways(s.substr(0, i), move(memo)))
          for (const int b : ways(s.substr(i + 1), move(memo)))
            if (s[i] == '+')
              ans.push_back(a + b);
            else if (s[i] == '-')
              ans.push_back(a - b);
            else
              ans.push_back(a * b);

    return memo[s] = (ans.empty() ? vector<int>{stoi(s)} : ans);
  }
};

Java

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

  private List<Integer> ways(final String s, Map<String, List<Integer>> memo) {
    if (memo.containsKey(s))
      return memo.get(s);

    List<Integer> ans = new ArrayList<>();

    for (int i = 0; i < s.length(); ++i)
      if (!Character.isDigit(s.charAt(i)))
        for (final int a : ways(s.substring(0, i), memo))
          for (final int b : ways(s.substring(i + 1), memo))
            if (s.charAt(i) == '+')
              ans.add(a + b);
            else if (s.charAt(i) == '-')
              ans.add(a - b);
            else
              ans.add(a * b);

    if (ans.isEmpty()) { // single number
      memo.put(s, Arrays.asList(Integer.parseInt(s)));
      return memo.get(s);
    }
    memo.put(s, ans);
    return ans;
  }
}

Python

class Solution:
  @lru_cache(None)
  def diffWaysToCompute(self, expression: str) -> List[int]:
    ans = []

    for i, c in enumerate(expression):
      if c in '+-*':
        for a in self.diffWaysToCompute(expression[:i]):
          for b in self.diffWaysToCompute(expression[i + 1:]):
            ans.append(eval(str(a) + c + str(b)))

    return ans or [int(expression)]

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