<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:
    // Function to calculate the maximum gain by removing "ab" or "ba" substrings
    int maximumGain(string s, int x, int y) {
        // If x is less than y, reverse the string and swap values of x and y to 
        // prioritize removing "ba" before "ab" since we want to maximize the gain
        if (x < y) {
            reverse(s.begin(), s.end());
            return maximumGain(s, y, x);
        }
        // Variable to store the total gain
        int totalGain = 0;
        stack<char> stkAb; // Stack to manage "ab" pairs
        stack<char> stkBa; // Stack to manage "ba" pairs

        // Iterate through the characters of the string
        for (char c : s) {
            // If the current character is 'a' or it's not 'b', push it onto stkAb
            if (c != 'b')
                stkAb.push(c);
            else {
                // If the top of stkAb is 'a', we found "ab" so we pop 'a' and 
                // add x to totalGain
                if (!stkAb.empty() && stkAb.top() == 'a') {
                    stkAb.pop();
                    totalGain += x;
                } else {
                    // Otherwise, push 'b' to stkAb
                    stkAb.push(c);
                }
            }
        }

        // While stkAb is not empty, we will manage "ba" pairs
        while (!stkAb.empty()) {
            char c = stkAb.top();
            stkAb.pop();
            // If the current character is 'a' or it's not 'b', push it onto stkBa
            if (c != 'b')
                stkBa.push(c);
            else {
                // If the top of stkBa is 'a', we found "ba" so we pop 'a' and 
                // add y to totalGain
                if (!stkBa.empty() && stkBa.top() == 'a') {
                    stkBa.pop();
                    totalGain += y;
                } else {
                    // Otherwise, push 'b' to stkBa
                    stkBa.push(c);
                }
            }
        }

        // Return the calculated total gain
        return totalGain;
    }
};

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