<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> diffWaysToCompute(string s) {
        
        vector< pair<char, int> > arr;

        int n = s.length();

        for( int i=0; i<n; i++) {
            if((s[i]=='+') || (s[i]=='-') || (s[i]=='*')) {
                arr.push_back({ s[i], 0});
            } else {
                int j = i;
                while((j+1<n) && isdigit(s[j+1])) {
                    j += 1;
                }
                int v = stoi(s.substr( i, j-i+1));
                i=j;
                arr.push_back({ '$', v});
            }
        }

        int m = arr.size();

        vector< vector< vector<int> > > dp( m, vector< vector<int> >( m, vector<int>()));

        auto eval = []( char op, int lval, int rval)->int {
            if(op=='+') {
                return lval + rval;
            } else if(op=='-') {
                return lval - rval;
            } else {
                return lval*rval;
            }
        };

        function<void(int,int)> solve;

        solve = [&]( int sid, int eid)->void {
            if(dp[sid][eid].empty()) {
                if(sid==eid) {
                    dp[sid][eid].push_back(arr[sid].second);
                } else {
                    for( int i=sid+1; i<=eid; i+=2) {
                        solve( sid, i-1);
                        solve( i+1, eid);
                        char op = arr[i].first;
                        for( int lv: dp[sid][i-1]) {
                            for( int rv: dp[i+1][eid]) {
                                dp[sid][eid].push_back(eval(op, lv, rv));
                            }
                        }
                    }
                }
            }
        };

        solve( 0, m-1);

        vector<int> res;
        for( int val: dp[0][m-1]) {
            res.push_back(val);
        }

        return res;
    }
};
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