<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 expression) {
        int n = expression.size();
        vector<vector<vector<int>>> memo(n, vector<vector<int>>(n));
        // memo[i][j] means from i-th to j-th char, list all different kind of 
        // value it can come up to.
        dfsMemo(expression, 0, n-1, memo);
        return memo[0][n-1];
    }
    
    // save the result to memo is enough, do not return vector<int>
    void dfsMemo(const string& exp, int i, int j, 
                 vector<vector<vector<int>>>& memo) {
        // cout << i << " " << j << endl;
        if(i>j) return; // nothing to split.
        // if(!memo[i][j].empty()) return;
        vector<int> res;
        // case1: split the expression with parenthesis.
        for(int k = i; k<=j; k++) {
            char c = exp.at(k);
            if(c=='+' || c=='-' || c=='*') {
                dfsMemo(exp, i, k-1, memo);
                dfsMemo(exp, k+1, j, memo);
                for(int x : memo[i][k-1]) {
                    for(int y : memo[k+1][j]) {
                        if(c=='+') res.push_back(x+y);
                        else if(c=='-') res.push_back(x-y);
                        else if(c=='*') res.push_back(x*y);
                    }
                }
            }
        }
        if(res.empty()) {
            // which means there is no + - *
            int num = 0;
            for(int k = i; k<=j; k++) {
                num = num*10 + (exp.at(k)-'0');
            }
            res.push_back(num);
        }
        memo[i][j] = move(res);
    }
};
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