<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 evaluate(int a, int b, char op) {
    switch (op) {
        case '+': return a + b;
        case '-': return a - b;
        case '*': return a * b;
        case '/': return a / b;  // Integer division
        default: return 0;
    }
}
    vector<int> diffWaysToCompute(string expression) {
    vector<int> results;
    
    // Traverse through each character in the expression
    for (int i = 0; i < expression.size(); ++i) {
        char c = expression[i];
        
        // If the character is an operator
        if (c == '+' || c == '-' || c == '*' || c == '/') {
            // Divide the expression into two parts
            string left = expression.substr(0, i);
            string right = expression.substr(i + 1);
            
            // Recursively compute all results for the left and right parts
            vector<int> leftResults = diffWaysToCompute(left);
            vector<int> rightResults = diffWaysToCompute(right);
            
            // Combine the results from left and right using the current operator
            for (int l : leftResults) {
                for (int r : rightResults) {
                    results.push_back(evaluate(l, r, c));
                }
            }
        }
    }
    
    // If there were no operators, the expression is just a number
    if (results.empty()) {
        results.push_back(stoi(expression));
    }
    
    return results;
}


};
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