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1717. Maximum Score From Removing Substrings

Description

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

  • 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.

Solutions

  • class Solution {
    public int maximumGain(String s, int x, int y) {
        if (x < y) {
            return maximumGain(new StringBuilder(s).reverse().toString(), y, x);
        }
        int ans = 0;
        Deque<Character> stk1 = new ArrayDeque<>();
        Deque<Character> stk2 = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c != 'b') {
                stk1.push(c);
            } else {
                if (!stk1.isEmpty() && stk1.peek() == 'a') {
                    stk1.pop();
                    ans += x;
                } else {
                    stk1.push(c);
                }
            }
        }
        while (!stk1.isEmpty()) {
            char c = stk1.pop();
            if (c != 'b') {
                stk2.push(c);
            } else {
                if (!stk2.isEmpty() && stk2.peek() == 'a') {
                    stk2.pop();
                    ans += y;
                } else {
                    stk2.push(c);
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int maximumGain(string s, int x, int y) {
        if (x < y) {
            reverse(s.begin(), s.end());
            return maximumGain(s, y, x);
        }
        int ans = 0;
        stack<char> stk1;
        stack<char> stk2;
        for (char c : s) {
            if (c != 'b')
                stk1.push(c);
            else {
                if (!stk1.empty() && stk1.top() == 'a') {
                    stk1.pop();
                    ans += x;
                } else
                    stk1.push(c);
            }
        }
        while (!stk1.empty()) {
            char c = stk1.top();
            stk1.pop();
            if (c != 'b')
                stk2.push(c);
            else {
                if (!stk2.empty() && stk2.top() == 'a') {
                    stk2.pop();
                    ans += y;
                } else
                    stk2.push(c);
            }
        }
        return ans;
    }
};

  • class Solution:
    def maximumGain(self, s: str, x: int, y: int) -> int:
        if x < y:
            return self.maximumGain(s[::-1], y, x)
        ans = 0
        stk1, stk2 = [], []
        for c in s:
            if c != 'b':
                stk1.append(c)
            else:
                if stk1 and stk1[-1] == 'a':
                    stk1.pop()
                    ans += x
                else:
                    stk1.append(c)
        while stk1:
            c = stk1.pop()
            if c != 'b':
                stk2.append(c)
            else:
                if stk2 and stk2[-1] == 'a':
                    stk2.pop()
                    ans += y
                else:
                    stk2.append(c)
        return ans


  • func maximumGain(s string, x int, y int) int {
	if x < y {
		t := []rune(s)
		for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
			t[i], t[j] = t[j], t[i]
		}
		return maximumGain(string(t), y, x)
	}
	ans := 0
	var stk1 []byte
	var stk2 []byte
	for i := range s {
		if s[i] != 'b' {
			stk1 = append(stk1, s[i])
		} else {
			if len(stk1) > 0 && stk1[len(stk1)-1] == 'a' {
				stk1 = stk1[0 : len(stk1)-1]
				ans += x
			} else {
				stk1 = append(stk1, s[i])
			}
		}
	}
	for _, c := range stk1 {
		if c != 'a' {
			stk2 = append(stk2, c)
		} else {
			if len(stk2) > 0 && stk2[len(stk2)-1] == 'b' {
				stk2 = stk2[0 : len(stk2)-1]
				ans += y
			} else {
				stk2 = append(stk2, c)
			}
		}
	}
	return ans
}

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