<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<pair<int,int>,vector<int>>dp;
    vector<int> solve(int i,int j,string &s){
        if(j-i+1<=2){
            string temp=s.substr(i,(j-i+1));
            cout<<temp<<endl;
            return {stoi(temp)};
        }

        if(dp.count({i,j})>0) return dp[{i,j}];

        vector<int>ans;

        for(int k=i;k<=j;k++){
            if(!isdigit(s[k])){
            vector<int> temp=solve(i,k-1,s);
            vector<int> tmp=solve(k+1,j,s);

                if(s[k]=='+'){
                    for(int it=0;it<temp.size();it++){
                        for(int itt=0;itt<tmp.size();itt++){
                            ans.push_back(temp[it]+tmp[itt]);
                        }
                    }
                }
                else if(s[k]=='*'){
                    for(int it=0;it<temp.size();it++){
                        for(int itt=0;itt<tmp.size();itt++){
                            ans.push_back(temp[it]*tmp[itt]);
                        }
                    }
                }
                else if(s[k]=='-'){
                    for(int it=0;it<temp.size();it++){
                        for(int itt=0;itt<tmp.size();itt++){
                            ans.push_back(temp[it]-tmp[itt]);
                        }
                    }
                }
            }
        }
        return dp[{i,j}]=ans;
    }
    vector<int> diffWaysToCompute(string s) {
        dp.clear();
        return solve(0,s.size()-1,s);
    }
};
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