<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:
    // Memoization map to store already computed results
    unordered_map<string, vector<int>> map;

    vector<int> diffWaysToCompute(string expression) {
        // Step 1: Check if the result for the current expression is already computed
        if (map.find(expression) != map.end()) {
            return map[expression];  // Return cached result
        }

        vector<int> result;

        // Step 2: Traverse the expression to find operators (+, -, *)
        for (int i = 0; i < expression.length(); i++) {
            char ch = expression[i];

            // Step 3: If the character is an operator, split the expression into left and right parts
            if (ch == '*' || ch == '+' || ch == '-') {
                // Recursive call to compute the result for the left sub-expression
                vector<int> left = diffWaysToCompute(expression.substr(0, i));
                
                // Recursive call to compute the result for the right sub-expression
                vector<int> right = diffWaysToCompute(expression.substr(i + 1));

                // Step 4: Combine the results from left and right sub-expressions
                for (int l : left) {
                    for (int r : right) {
                        if (ch == '+') {
                            result.push_back(l + r);  // Add the results
                        } else if (ch == '-') {
                            result.push_back(l - r);  // Subtract the results
                        } else if (ch == '*') {
                            result.push_back(l * r);  // Multiply the results
                        }
                    }
                }
            }
        }

        // Step 5: Base case - If there were no operators, it means the expression is a single number
        if (result.empty()) {
            result.push_back(stoi(expression));  // Convert string to integer
        }

        // Step 6: Store the result in the map to avoid redundant computation
        map[expression] = result;

        // Step 7: Return the final computed result
        return result;
    }
};
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