<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:
    int func(int f,int s, char op){
        if(op=='+'){
            return f+s;
        }
        if(op=='-'){
            return f-s;
        }
        if(op=='*'){
            return f*s;
        }
        return 0;
    }
    vector<int> diffWaysToCompute(string s) {
        vector<int>ans;
        int n=s.size();
        vector<int>num;
        vector<char>op;
        string st="";
        for(int i=0;i<n;i++){
            if(s[i]=='+' || s[i]=='-' || s[i]=='*'){
                int x=stoi(st);
                num.push_back(x);
                st="";
                op.push_back(s[i]);
            }else{
                st+=s[i];
            }
        }
        if(st!=""){
            int x=stoi(st);
            num.push_back(x);
        }
        n=num.size();
        // vector<vector<int>>dp(n,vector<int>(n,0));
        vector<int> dp[n][n];
        for(int i=0;i<n;i++){
            dp[i][i].push_back(num[i]);
        }
        for(int len=2;len<=n;len++){
            for(int i=0;i<n;i++){
                int j=i+len-1;
                if(j>=n) continue;
                for(int k=i;k<j;k++){
                    char opera=op[k];
                    for(auto &f:dp[i][k]){
                        for(auto &s:dp[k+1][j]){
                            int result=func(f,s,opera);
                            dp[i][j].push_back(result);
                        }
                    }
                    // if(len==n){
                    // ans.push_back(result);
                    // cout<<len<<" "<<i<<" "<<k<<" "<<j<<" "<<result<<endl;
                    // }
                }
            }
        }
        for(auto &it:dp[0][n-1]){
            ans.push_back(it);
        }
        return ans;
    }
};
8000