<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:
    //dp[index1][index2]
    vector<int> solve(int index1, int index2, string expression, vector<vector<vector<int>>>& dp) {
        //base case
        if(index1 == index2) {
            return dp[index1][index2] = {expression[index1] - '0'};
        }
        if(index1 + 1 == index2) {
            return dp[index1][index2] = {(expression[index1] - '0')*10 + (expression[index2] - '0')};
        }
        if(dp[index1][index2].size() != 0) return dp[index1][index2];
        vector<int> v;
        for(int i = index1; i <= index2; i++) {
            if(expression[i] == '*' or expression[i] == '-' or expression[i] == '+') {
                vector<int> temp = solve(index1, i - 1, expression, dp);
                vector<int> temp1 = solve(i + 1, index2, expression, dp);
                for(int j = 0; j < temp.size(); j++) {
                    for(int k = 0; k < temp1.size(); k++) {
                        if(expression[i] == '*') {
                            v.push_back(temp[j]*temp1[k]);
                        }
                        else if(expression[i] == '+') {
                            v.push_back(temp[j] + temp1[k]);
                        }
                        else {
                            v.push_back(temp[j] - temp1[k]);
                        }
                    }
                }
            } 
        }
        return dp[index1][index2] = v;
    }
    vector<int> diffWaysToCompute(string expression) {
        vector<vector<vector<int>>> dp(expression.size(), vector<vector<int>>(expression.size(), vector<int>()));
        return solve(0, expression.size() - 1, expression, dp);
    }
};
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