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 {
    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;
    }
}