<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:
    int compute(int a,int b,char c){
        if(c=='*')
          return a*b;
        if(c=='+')
          return a+b;
        return a-b;
    }
    bool isOperator(char ch){
        if(ch=='+'||ch=='-'||ch=='*')
          return true;
        return false;
    }
    vector<int> diffWaysToCompute(string expression) {
        int n=expression.length();
        vector<vector<vector<int>>> dp(n+1,vector<vector<int>>(n+1));
        for(int i=0;i<n;i++){
            if(!isOperator(expression[i]))
               dp[i][i]={expression[i]-'0'};
        }
        for(int i=0;i<n-1;i++){
            if((i==0||isOperator(expression[i-1])) && (i==n-2 || isOperator(expression[i+2])))
               dp[i][i+1]={stoi(expression.substr(i,2))};
        }
        for(int l=3;l<=n;l++){
            for(int i=0;i<=n-l;i++){
                for(int j=i;j<l+i;j++){
                    if(isOperator(expression[j])){
                        vector<int> left=dp[i][j-1];
                        vector<int> right=dp[j+1][l+i-1];
                        for(int t1:left){
                            for(int t2:right){
                                dp[i][l+i-1].push_back(compute(t1,t2,expression[j]));
                            }
                        }
                    }
                }
            }
        }
        return dp[0][n-1];
    }
};
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