<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:
    map<pair<int,int>, vector<int>> dp;
    int calculate(int x, int y, char opr){
        int ans;
        switch(opr){
            case '+':
                ans = x + y;
                break;
            case '-':
                ans = x - y;
                break;
            case '*':
                ans = x * y;
                break;
            default:
                break;
        }
        return ans;
    }
    vector<int> rec(string &expression, int i, int j){
        if(dp.count({i,j})) return dp[{i,j}];

        bool isNumber = true;
        vector<int> ans;
        for(int k = i; k < j; k++){
            if(!isdigit(expression[k])){
                vector<int> left = rec(expression, i, k-1);
                vector<int> right = rec(expression, k+1, j);

                for(auto x : left){
                    for(auto y : right){
                        ans.push_back(calculate(x, y, expression[k]));
                    }
                }

                isNumber = false;
            }
        }
        if(isNumber){
            ans.push_back(stoi(expression.substr(i,j-i+1)));
        }
        return dp[{i,j}] = ans;
    }
    vector<int> diffWaysToCompute(string expression) {
        int n = expression.size();
        return rec(expression, 0, n-1);
    }
};
9100