<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>

vector<int> f(string &s,int l,int r)
{
    bool t=false;
    vector<int> left;
    vector<int> right;
    vector<int> ans;
    if(l==r) 
    { 
        ans.push_back(s[l]-'0');
        return ans;
    }
    //if(dp[l][r]!=-1)return dp[l][r];
    int a,b;
    for(int k=l;k<r;k++)
    {
        if(s[k]=='*')
        {
            t=true;
            left=f(s,l,k-1);
            right=f(s,k+1,r);
            int nl=left.size();
            int nr=right.size();
           for(int i=0;i<nl;i++)
            {
                for(int j=0;j<nr;j++)
                {
                    
                    ans.push_back(left[i]*right[j]);
                }
            }
        }
        if(s[k]=='-')
        {
            t=true;
            left=f(s,l,k-1);
            right=f(s,k+1,r);
            int nl=left.size();
            int nr=right.size();
           for(int i=0;i<nl;i++)
            {
                for(int j=0;j<nr;j++)
                {
                    
                   ans.push_back(left[i]-right[j]);
                }
            }
        }
        if(s[k]=='+')
        {
            t=true;
            left=f(s,l,k-1);
            right=f(s,k+1,r);
            int nl=left.size();
            int nr=right.size();
           for(int i=0;i<nl;i++)
            {
                for(int j=0;j<nr;j++)
                {
                    
                    ans.push_back(left[i]+right[j]);
                }
            }
        }

    }
    int temp=0;
    if(!t)
    {
        for(int i=l;i<=r;i++)
        temp=temp*10+(s[i]-'0');
        ans.push_back(temp);
    }
    return ans;
}
class Solution {
public:
    vector<int> diffWaysToCompute(string expression) {
        vector<int> ans;
        //vector<vector<int>> dp(expression.size(),vector<int>(expression.size(),-1));
        
        return f(expression,0,expression.size()-1);
        
    }
};
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