<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 op(int i,int j, char opcode)
    {
        if(opcode == '+')
            return i + j;
        if(opcode == '-')
            return i - j;
        if(opcode == '*')
            return i * j;
        return -1;
    }
    vector<int> solve(string& exp, int i, int j)
    {   
        // if(i == j)
        //     return {exp[i]-'0'};
        int currNum = 0, f = 0;
        vector<int> v;
        for(int k=i;k<=j;k++)
        {
            if(exp[k] >= '0' && exp[k] <= '9')
                currNum = currNum*10 + exp[k]-'0';
            else
            {
                vector<int> l = solve(exp, i, k-1);
                vector<int> r = solve(exp, k+1, j);
                for(auto x : l)
                {
                    for(auto y : r)
                    {
                        int res = op(x,y,exp[k]);
                        v.push_back(res);
                    }
                }
                currNum = 0;
                f=1;
            }
        }
        if(!f)
            return {currNum};
        return v;
    }
    vector<int> diffWaysToCompute(string expression) {
        return solve(expression, 0, expression.size()-1);      
    }
};
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