<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> results;

        // Base case: if the string is empty, return an empty list
        if (expression.length() == 0) return results;

        // Base case: if the string is a single character, treat it as a number
        // and return it
        if (expression.length() == 1) {
            results.push_back(stoi(expression));
            return results;
        }

        // If the string has only two characters and the first character is a
        // digit, parse it as a number
        if (expression.length() == 2 && isdigit(expression[0])) {
            results.push_back(stoi(expression));
            return results;
        }

        // Recursive case: iterate through each character
        for (int i = 0; i < expression.length(); i++) {
            char currentChar = expression[i];

            // Skip if the current character is a digit
            if (isdigit(currentChar)) continue;

            // Split the expression into left and right parts
            vector<int> leftResults =
                diffWaysToCompute(expression.substr(0, i));
            vector<int> rightResults =
                diffWaysToCompute(expression.substr(i + 1));

            // Combine results from left and right parts
            for (int leftValue : leftResults) {
                for (int rightValue : rightResults) {
                    int computedResult = 0;

                    // Perform the operation based on the current character
                    switch (currentChar) {
                        case '+':
                            computedResult = leftValue + rightValue;
                            break;
                        case '-':
                            computedResult = leftValue - rightValue;
                            break;
                        case '*':
                            computedResult = leftValue * rightValue;
                            break;
                    }

                    results.push_back(computedResult);
                }
            }
        }

        return results;
    }
};
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