Leetcode 1717. Maximum Score From Removing Substrings

2024/07/12 的每日一題！

題目描述

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.

範例

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 <= 10^5
• 1 <= x, y <= 10^4
• s consists of lowercase English letters.

解法

方法：寫一個叫做 remove 的方法，該方法利用 stack 消除所有 removestr，並將每個 removestr 作為 val 值加總在一起並回傳。 針對題目字串，如果ｘ較大，我們就先處理 ab，如 y 較大，處理 ba。

func maximumGain(s string, x int, y int) int {
    
    remove := func(str string, removestr string, val int) int {
        count := 0
        stack := make([]byte, 0)

        for i:=0; i<len(str); i++ {
            stack = append(stack, str[i])

            l := len(stack)
            if l >= 2 && stack[l-1]==removestr[1] && stack[l-2]==removestr[0] {
                count += val
                stack = stack[:l-2]
            }
        }

        s = string(stack)

        return count
    }

    if x > y { 
        return remove(s, "ab", x) + remove(s, "ba", y) 
    } else {
        return remove(s, "ba", y) + remove(s, "ab", x)
    }
    

    return 0
}

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