<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> dp[20][20];
    vector<int> convert(string& expression){
        vector<int> ans;
        int sz=expression.size();
        ans.reserve(sz);
        int x=0;
        for(int i=0; i<sz; i++){
            const char c=expression[i];
            switch(c){
                case '+': ans.push_back(101); break;// 101 for +
                case '-': ans.push_back(102); break;// 102 for -
                case '*': ans.push_back(103); break;// 103 for *
                default:{
                    x=10*x+c-'0';
                    if (i==sz-1 || !isdigit(expression[i+1])){
                        ans.push_back(x);
                        x=0;
                    }
                }
            }
        }
        return ans;
    }

    vector<int> f(int l, int r, vector<int>& nums){
        if (l>r) return {};
        if (l==r){
            if (nums[l]>=101) return {};
            return {nums[l]};
        }
        if (dp[l][r].size()>0) return dp[l][r];

        vector<int> ans;
        for(int i=l; i<=r; i++){
            if (nums[i]<100) continue;
            vector<int> Left=f(l, i-1, nums);
            vector<int> Right=f(i+1, r, nums);
            for(int L: Left)
                for(int R: Right){
                    int x=0;
                    switch(nums[i]){
                        case 101: ans.push_back(L+R); break;
                        case 102: ans.push_back(L-R); break;
                        case 103: ans.push_back(L*R); break;
                    }
                }
        }
        return dp[l][r]=ans;
    }

    vector<int> diffWaysToCompute(string& expression) {
        vector<int> nums=convert(expression);
        int n=nums.size();
        return f(0, n-1, nums);
    }
};
9000