<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>ans;
        unordered_map<string, vector<int>>memo;

        for(int ind=0;ind<expression.length();ind++)
        {

            vector<int>left, right;
            char cur= expression[ind];

            if(cur=='*' || cur == '-' || cur=='+')
            {
                string e1 = expression.substr(0,ind);
                string e2 = expression.substr(ind+1);
                if(memo.find(e1)!=memo.end())
                {
                    left=memo[e1];
                }
                else {
                    left=diffWaysToCompute(e1);
                }
                if(memo.find(e2)!=memo.end())
                {
                    right=memo[e2];
                }
                else{
                    right=diffWaysToCompute(e2);
                }
                for(auto i:left)
                {
                    for(auto j:right)
                    {
                        if(cur=='*') 
                        {
                            ans.push_back(i*j);
                        }
                        else if(cur=='+')
                        {
                            ans.push_back(i+j);
                        }
                        else{
                            ans.push_back(i-j);
                        }
                    }
                }
            }
        }
        if(ans.empty()) ans.push_back(stoi(expression));
        return ans;
    }
};
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