Different Ways to Add Parentheses
<p>Given a string <code>expression</code> of numbers and operators, return <em>all possible results from computing all the different possible ways to group numbers and operators</em>. You may return the answer in <strong>any order</strong>.</p>

<p>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 <code>10<sup>4</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> expression = &quot;2-1-1&quot;
<strong>Output:</strong> [0,2]
<strong>Explanation:</strong>
((2-1)-1) = 0 
(2-(1-1)) = 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> expression = &quot;2*3-4*5&quot;
<strong>Output:</strong> [-34,-14,-10,-10,10]
<strong>Explanation:</strong>
(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
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= expression.length &lt;= 20</code></li>
	<li><code>expression</code> consists of digits and the operator <code>&#39;+&#39;</code>, <code>&#39;-&#39;</code>, and <code>&#39;*&#39;</code>.</li>
	<li>All the integer values in the input expression are in the range <code>[0, 99]</code>.</li>
	<li>The integer values in the input expression do not have a leading <code>&#39;-&#39;</code> or <code>&#39;+&#39;</code> denoting the sign.</li>
</ul>
Recursion; Memoization; Math; String; Dynamic Programming
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 == '+' || c == '*') {
                vector<int> left = dfs(exp.substr(0, i));
                vector<int> right = dfs(exp.substr(i + 1, n - i - 1));
                for (int& a : left) {
                    for (int& b : right) {
                        if (c == '-')
                            ans.push_back(a - b);
                        else if (c == '+')
                            ans.push_back(a + b);
                        else
                            ans.push_back(a * b);
                    }
                }
            }
        }
        memo[exp] = ans;
        return ans;
    }

private:
    unordered_map<string, vector<int>> memo;
};