<p>You are given a string <code>s</code> and two integers <code>x</code> and <code>y</code>. You can perform two types of operations any number of times.</p>

<ul>
	<li>Remove substring <code>&quot;ab&quot;</code> and gain <code>x</code> points.

	<ul>
		<li>For example, when removing <code>&quot;ab&quot;</code> from <code>&quot;c<u>ab</u>xbae&quot;</code> it becomes <code>&quot;cxbae&quot;</code>.</li>
	</ul>
	</li>
	<li>Remove substring <code>&quot;ba&quot;</code> and gain <code>y</code> points.
	<ul>
		<li>For example, when removing <code>&quot;ba&quot;</code> from <code>&quot;cabx<u>ba</u>e&quot;</code> it becomes <code>&quot;cabxe&quot;</code>.</li>
	</ul>
	</li>
</ul>

<p>Return <em>the maximum points you can gain after applying the above operations on</em> <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cdbcbbaaabab&quot;, x = 4, y = 5
<strong>Output:</strong> 19
<strong>Explanation:</strong>
- Remove the &quot;ba&quot; underlined in &quot;cdbcbbaaa<u>ba</u>b&quot;. Now, s = &quot;cdbcbbaaab&quot; and 5 points are added to the score.
- Remove the &quot;ab&quot; underlined in &quot;cdbcbbaa<u>ab</u>&quot;. Now, s = &quot;cdbcbbaa&quot; and 4 points are added to the score.
- Remove the &quot;ba&quot; underlined in &quot;cdbcb<u>ba</u>a&quot;. Now, s = &quot;cdbcba&quot; and 5 points are added to the score.
- Remove the &quot;ba&quot; underlined in &quot;cdbc<u>ba</u>&quot;. Now, s = &quot;cdbc&quot; and 5 points are added to the score.
Total score = 5 + 4 + 5 + 5 = 19.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aabbaaxybbaabb&quot;, x = 5, y = 4
<strong>Output:</strong> 20
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= x, y &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    int maximumGain(string s, int x, int y) {
        std::string expensive, cheap;

        if(x >= y){
            expensive = "ab";
            cheap = "ba";
        }
        else{
            expensive = "ba";
            cheap = "ab";
            swap(x,y);
        }

        int res = 0;
        stack<char> temp1;
        for(char c: s){
            if(!temp1.empty() && temp1.top() == expensive[0] && c == expensive[1]){
                res += x;
                temp1.pop();
            }
            else{
                temp1.push(c);
            }
        }

        vector<char> remaining;
        while (!temp1.empty()) {
            remaining.push_back(temp1.top());
            temp1.pop();
        }

        stack<char> temp2;
        // 逆序遍历vector以模拟原始栈顺序
        for (auto it = remaining.rbegin(); it != remaining.rend(); ++it) {
            char c = *it;
            if (!temp2.empty() && temp2.top() == cheap[0] && c == cheap[1]) {
                res += y;
                temp2.pop();
            } else {
                temp2.push(c);
            }
        }
        return res;
    }
};



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