<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:
    // greedy
    // time: O(n)
    // space: O(1)
    // b(ab)a: Assume score(ab) > score(ba) but removing ba first is optimal.
    // But after removing ab we can remove one ba, so we get score(ab) + score(ba) which is greater than score(ba) * 2.
    int maximumGain(string s, int x, int y) {
        string p1 = "ab", p2 = "ba";
        if (x < y) {
            swap(p1, p2);
            swap(x, y);
        }
        return helper(s, p1, x) + helper(s, p2, y);
    }

    int helper(string& s, string pattern, int score) {
        int index = 0;
        int res = 0;
        for (int i = 0; i < s.size(); ++i) {
            s[index++] = s[i];
            if (index > 1 && s.substr(index-2, 2) == pattern) {
                index -=  2;
                res += score;
            }
        }
        s.resize(index);
        return res;
    }
};
19259