<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) {
        vector<int> nums;
        vector<char> op;
        int size = expression.size();
        for(int i=0;i<size;) {
            int c = 0;
            while(i<size && !(expression[i]=='-' || expression[i]=='+' || expression[i]=='*')) {
                c = c*10 + expression[i]-'0';
                i++;
            }
            nums.push_back(c);
            if(i<size)
            op.push_back(expression[i++]);
        }
        int n = nums.size();
        vector<int> dp[n][n];
        for(int i=0;i<n;i++)
        dp[i][i].push_back(nums[i]);
        for(int cl=1;cl<n;cl++) {
            int i = 0;
            int j = cl;
            while(j<n) {
                for(int k=i;k<j;k++) {
                    for(auto &x:dp[i][k]) {
                        for(auto &y:dp[k+1][j]) {
                            if(op[k]=='-')
                            dp[i][j].push_back(x-y);
                            else if(op[k]=='+')
                            dp[i][j].push_back(x+y);
                            else
                            dp[i][j].push_back(x*y);
                        }
                    }
                }
                i++;
                j++;
            }
        }
        return dp[0][n-1];
    }
};
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