<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:
    map<string,vector<int>> dp;
    vector<int> func(string q){
        if(dp.find(q)!=dp.end())
            return dp[q];      

        vector<int> ans;
        for(int i=0;i<q.size();i++){
            if(!(q[i]=='+' || q[i]=='-' || q[i]=='*'))
                continue;
            string q1=q.substr(0,i);
            string q2=q.substr(i+1);
            vector<int> v1=func(q1);
            vector<int> v2=func(q2);

            for(auto &j:v1){
                for(auto &k:v2){
                    if(q[i]=='+')
                        ans.push_back(j+k);
                    else if(q[i]=='-')
                        ans.push_back(j-k);
                    else
                        ans.push_back(j*k);
                }
            }                                            
        }
        if(ans.empty()){ 
            int no=stoi(q);
            return dp[q]= {no};
        }
        return dp[q]= ans;
        
    }
    vector<int> diffWaysToCompute(string q) {
        dp.clear();
        return func(q);        
    }
};
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