<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>>> memoization(n+1, vector<vector<int>>(n+1, vector<int>()));
        diffWaysToCompute(expression, 0, n, memoization);
        return memoization[0][n];
    }
    void diffWaysToCompute(string& expression, int low, int high, vector<vector<vector<int>>>& memoization) {
        for(int i = low; i < high; ++i) {
            char c = expression[i];
            if(c == '+' || c == '-' || c == '*') {
                if(memoization[low][i].empty()) 
                    diffWaysToCompute(expression, low, i, memoization);
                if(memoization[i+1][high].empty())
                    diffWaysToCompute(expression, i+1, high, memoization);
                int val = 0;
                for(auto l : memoization[low][i]) {
                    for(auto r : memoization[i+1][high]) {
                        switch(c) {
                            case '+' : val = l + r; break;
                            case '-' : val = l - r; break;
                            case '*' : val = l * r; break;
                        }
                        memoization[low][high].push_back(val);
                    }
                }
            }
        }
        if (memoization[low][high].empty()) 
            memoization[low][high].push_back(stoi(expression.substr(low, high)));

    }
};
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