<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>
</ul>

class Solution {
public:

    string s;
    int op(char c, int a, int b){
        if(c == '+'){
            return a+b;
        }else if(c == '-'){
            return a-b;
        }
        return a*b;
    }

    vector<int> rec(int l, int r){
        // if(l == r){
        //     int x = s[l]-'0';
        //     return {x};
        // }
        bool isNum = true;
        int number = 0;
        for(int i = l; i <= r; i++){
            if(!isdigit(s[i])){
                isNum = false;
                break;
            }
            number = number*10+(s[i]-'0');
        }
        if(isNum) return {number};
        vector<int> ans;
        for(int i = l; i <= r; i++){
            if(!isdigit(s[i])){
                vector<int> left = rec(l, i-1);
                vector<int> right = rec(i+1, r);
                for(auto x : left){
                    for(auto y : right){
                        ans.push_back(op(s[i], x, y));
                    }
                }
            }
        }
        return ans;
    } 

    vector<int> diffWaysToCompute(string expression) {
        this->s = expression;
        return rec(0, s.length()-1);
    }
};
12600