Different Ways to Add Parentheses LeetCode Solution

Problem – Different Ways to Add Parentheses LeetCode Solution

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].

Different Ways to Add Parentheses LeetCode Solution in Java

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> ret = new LinkedList<Integer>();
        for (int i=0; i<input.length(); i++) {
            if (input.charAt(i) == '-' ||
                input.charAt(i) == '*' ||
                input.charAt(i) == '+' ) {
                String part1 = input.substring(0, i);
                String part2 = input.substring(i+1);
                List<Integer> part1Ret = diffWaysToCompute(part1);
                List<Integer> part2Ret = diffWaysToCompute(part2);
                for (Integer p1 :   part1Ret) {
                    for (Integer p2 :   part2Ret) {
                        int c = 0;
                        switch (input.charAt(i)) {
                            case '+': c = p1+p2;
                                break;
                            case '-': c = p1-p2;
                                break;
                            case '*': c = p1*p2;
                                break;
                        }
                        ret.add(c);
                    }
                }
            }
        }
        if (ret.size() == 0) {
            ret.add(Integer.valueOf(input));
        }
        return ret;
    }
}

Different Ways to Add Parentheses LeetCode Solution in C++

class Solution {
public:
	vector<int> diffWaysToCompute(string input) {
		unordered_map<string, vector<int>> dpMap;
		return computeWithDP(input, dpMap);
	}

	vector<int> computeWithDP(string input, unordered_map<string, vector<int>> &dpMap) {
		vector<int> result;
		int size = input.size();
		for (int i = 0; i < size; i++) {
			char cur = input[i];
			if (cur == '+' || cur == '-' || cur == '*') {
				// Split input string into two parts and solve them recursively
				vector<int> result1, result2;
				string substr = input.substr(0, i);
				// check if dpMap has the result for substr
				if (dpMap.find(substr) != dpMap.end())
					result1 = dpMap[substr];
				else
					result1 = computeWithDP(substr, dpMap);

				substr = input.substr(i + 1);
				if (dpMap.find(substr) != dpMap.end())
					result2 = dpMap[substr];
				else
					result2 = computeWithDP(substr, dpMap);
				
				for (auto n1 : result1) {
					for (auto n2 : result2) {
						if (cur == '+')
							result.push_back(n1 + n2);
						else if (cur == '-')
							result.push_back(n1 - n2);
						else
							result.push_back(n1 * n2);
					}
				}
			}
		}
		// if the input string contains only number
		if (result.empty())
			result.push_back(atoi(input.c_str()));
		// save to dpMap
		dpMap[input] = result;
		return result;
	}
};

Different Ways to Add Parentheses LeetCode Solution in Python

class Solution(object):
    def diffWaysToCompute(self, input):
        m = {}
        return self.dfs(input, m)
        
    def dfs(self, input, m):
        if input in m:
            return m[input]
        if input.isdigit():
            m[input] = int(input)
            return [int(input)]
        ret = []
        for i, c in enumerate(input):
            if c in "+-*":
                l = self.diffWaysToCompute(input[:i])
                r = self.diffWaysToCompute(input[i+1:])
                ret.extend(eval(str(x)+c+str(y)) for x in l for y in r)
        m[input] = ret
        return ret

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