241. Different Ways To Add Parentheses¶

Difficulty: Medium

241. Different Ways to Add Parentheses

Medium

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.

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);
        ArrayList<Integer> values = new ArrayList<Integer>();
        if (!hasOperator(expression))
            values.add(Integer.parseInt(expression));
        else
            for (int i = 0; i < expression.length(); i++) {
                char symbol = expression.charAt(i);
                if (!Character.isDigit(symbol)) {
                    List<Integer> leftParts = diffWaysToCompute(expression.substring(0, i), map);
                    List<Integer> rightParts = diffWaysToCompute(expression.substring(i + 1), map);
                    for (int l : leftParts)
                        for (int r : rightParts)
                            switch (symbol) {
                                case '+' -> values.add(l + r);
                                case '-' -> values.add(l - r);
                                case '*' -> values.add(l * r);
                            }
                }
            }
        return map.compute(expression, (k,v) -> values);
    }
    private boolean hasOperator(String expression) {
        for (var i = 0; i < expression.length(); i++)
            switch (expression.charAt(i)) {
                case '+', '-', '*' -> {
                    return true;
                }
            }
        return false;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here