<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 exp;
unordered_map<int,unordered_map<int,vector<int>>> dp;
vector<int> solve(int st,int e){
    if(dp[st].find(e)!=dp[st].end())
    return dp[st][e];
    if(st==e){
        return dp[st][e]={exp[st]-'0'};

    }
    if(e-st==1){
        return dp[st][e]={stoi(exp.substr(st,2))};
    }

 vector<int> ans;
    for(int i=st;i<=e;i++){
        if(exp[i]=='+'){
           vector<int> l=solve(st,i-1),r=solve(i+1,e);
           for(auto i:l){
                    for(auto j:r){
                        ans.push_back(i+j);
                    }
                }
        }
        else if(exp[i]=='-'){
            vector<int> l=solve(st,i-1),r=solve(i+1,e);
             for(auto i:l){
                    for(auto j:r){
                        ans.push_back(i-j);
                    }
                }
        }

        else if(exp[i]=='*'){
                vector<int> l=solve(st,i-1),r=solve(i+1,e);

               

                for(auto i:l){
                    for(auto j:r){
                        ans.push_back(i*j);
                    }
                }
        }
    }

    return dp[st][e]=ans;
}
    vector<int> diffWaysToCompute(string expression) {
        int st=0,e=expression.size()-1;
        this->exp=expression;
        return solve(0,e);
    }
};
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