• Remove substring "ab" and gain x points.
  • For example, when removing "ab" from "cabxbae" it becomes "cxbae".
• Remove substring "ba" and gain y points.
  • For example, when removing "ba" from "cabxbae" it becomes "cabxe".

Return the maximum points you can gain after applying the above operations on s.

Example 1:

Input: s = "cdbcbbaaabab", x = 4, y = 5
Output: 19
Explanation:
- Remove the "ba" underlined in "cdbcbbaaabab". Now, s = "cdbcbbaaab" and 5 points are added to the score.
- Remove the "ab" underlined in "cdbcbbaaab". Now, s = "cdbcbbaa" and 4 points are added to the score.
- Remove the "ba" underlined in "cdbcbbaa". Now, s = "cdbcba" and 5 points are added to the score.
- Remove the "ba" underlined in "cdbcba". Now, s = "cdbc" and 5 points are added to the score.
Total score = 5 + 4 + 5 + 5 = 19.

Example 2:

Input: s = "aabbaaxybbaabb", x = 5, y = 4
Output: 20

Constraints:

• 1 <= s.length <= 105
• 1 <= x, y <= 104
• s consists of lowercase English letters.

Solution: Greedy + Stack

Remove the pattern with the larger score first.

Using a stack to remove all occurrences of a pattern in place in O(n) Time.

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int maximumGain(string s, int x, int y) {
    // Remove patttern p from s for t points each.
    // Returns the total score.
    auto remove = [&](const string& p, int t) {
      int total = 0;
      int i = 0;
      for (int j = 0; j < s.size(); ++j, ++i) {
        s[i] = s[j];
        if (i && s[i - 1] == p[0] && s[i] == p[1]) {
          i -= 2;
          total += t;
        }
      }
      s.resize(i); 
      return total;
    };
    string px{"ab"};
    string py{"ba"};
    if (y > x) {
      swap(px, py);
      swap(x, y);
    }    
    return remove(px, x) + remove(py, y);
  }
};

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