<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:
    vector<int> solve(string& e,int l,int r){
        if(l>r){return {};}
        if(l==r){return {e[l]-'0'};}
        vector<int>ans;
        int f=0;
        for(int i=l;i<=r;i++){
            if(e[i]=='+'){
                f=1;
                vector<int>left=solve(e,l,i-1);
                vector<int>right=solve(e,i+1,r);
                for(int j=0;j<left.size();j++){
                    for(int k=0;k<right.size();k++){
                        ans.push_back(left[j]+right[k]);
                    }
                }
            }
            else if(e[i]=='-'){
                f=1;
                vector<int>left=solve(e,l,i-1);
                vector<int>right=solve(e,i+1,r);
                for(int j=0;j<left.size();j++){
                    for(int k=0;k<right.size();k++){
                        ans.push_back(left[j]-right[k]);
                    }
                }
            }
            else if(e[i]=='*'){
                f=1;
                vector<int>left=solve(e,l,i-1);
                vector<int>right=solve(e,i+1,r);
                for(int j=0;j<left.size();j++){
                    for(int k=0;k<right.size();k++){
                        ans.push_back(left[j]*right[k]);
                    }
                }
            }
        }
        if(f==0){
            int num=0;
            for(int i=l;i<=r;i++){
                num=num*10+(e[i]-'0');
            }
            return {num};
        }
        

        return ans;
    }
    vector<int> diffWaysToCompute(string expression) {
        
        return solve(expression,0,expression.size()-1);
    }
};
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