<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> fun(vector<int>&v,int s,int e){
        if(s==e)
            return {v[s]};
        if(s+2==e){
            if(v[(s+e)/2]==-1)
                return {v[s]+v[e]};
            else if(v[(s+e)/2]==-2)
                return {v[s]-v[e]};
            else if(v[(s+e)/2]==-3)
                return {v[s]*v[e]};
        }
        vector<int>ans;
        for(int j=s;j<=e-2;j+=2){
            vector<int>a=fun(v,s,j);
            vector<int>b=fun(v,j+2,e);
            for(int k=0;k<a.size();k++){
                for(int l=0;l<b.size();l++){
                    if(v[j+1]==-1)
                        ans.push_back(a[k]+b[l]);
                    else if(v[j+1]==-2)
                        ans.push_back(a[k]-b[l]);
                    else if(v[j+1]==-3)
                        ans.push_back(a[k]*b[l]);
                }
            }
        }
        return ans;
    }
    vector<int> diffWaysToCompute(string s) {
        vector<int>v;
        int i=0;
        while(i<s.length()){
            if(s[i]=='+'){
                v.push_back(-1);
                i++;
                continue;
            }
            if(s[i]=='-'){
                v.push_back(-2);
                i++;
                continue;
            }
            if(s[i]=='*'){
                v.push_back(-3);
                i++;
                continue;
            }
            int a=0;
            while(s[i]>='0' && s[i]<='9'){
                a=a*10+(s[i]-'0');
                i++;
            }
            v.push_back(a);
        }
        return fun(v,0,v.size()-1);
    }
};
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