<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) {
        // Initialize memoization vector to store results of subproblems
        vector<vector<vector<int>>> memo(
            expression.length(), vector<vector<int>>(expression.length()));
        // Solve for the entire expression
        return computeResults(expression, memo, 0, expression.length() - 1);
    }

private:
    vector<int> computeResults(string& expression,
                               vector<vector<vector<int>>>& memo, int start,
                               int end) {
        // If result is already memoized, return it
        if (!memo[start][end].empty()) {
            return memo[start][end];
        }

        vector<int> results;

        // Base case: single digit
        if (start == end) {
            results.push_back(expression[start] - '0');
            return results;
        }

        // Base case: two-digit number
        if (end - start == 1 && isdigit(expression[start])) {
            int tens = expression[start] - '0';
            int ones = expression[end] - '0';
            results.push_back(10 * tens + ones);
            return results;
        }

        // Recursive case: split the expression at each operator
        for (int i = start; i <= end; i++) {
            char currentChar = expression[i];
            if (isdigit(currentChar)) {
                continue;
            }

            vector<int> leftResults =
                computeResults(expression, memo, start, i - 1);
            vector<int> rightResults =
                computeResults(expression, memo, i + 1, end);

            // Combine results from left and right subexpressions
            for (int leftValue : leftResults) {
                for (int rightValue : rightResults) {
                    switch (currentChar) {
                        case '+':
                            results.push_back(leftValue + rightValue);
                            break;
                        case '-':
                            results.push_back(leftValue - rightValue);
                            break;
                        case '*':
                            results.push_back(leftValue * rightValue);
                            break;
                    }
                }
            }
        }

        // Memoize the result for this subproblem
        memo[start][end] = results;

        return results;
    }
};
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