23_Maximum_Score_From_Removing_Substrings.cpp

// 1717. Maximum Score From Removing Substrings
// Solved
// Medium
// Topics
// premium lock icon
// Companies
// Hint
// 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.

class Solution
{
public:
    int maximumGain(string s, int x, int y)
    {
        int aCount = 0;
        int bCount = 0;
        int lesser = min(x, y);
        int result = 0;

        for (char c : s)
        {
            if (c > 'b')
            {
                result += min(aCount, bCount) * lesser;
                aCount = 0;
                bCount = 0;
            }
            else if (c == 'a')
            {
                if (x < y && bCount > 0)
                {
                    bCount--;
                    result += y;
                }
                else
                {
                    aCount++;
                }
            }
            else
            {
                if (x > y && aCount > 0)
                {
                    aCount--;
                    result += x;
                }
                else
                {
                    bCount++;
                }
            }
        }

        result += min(aCount, bCount) * lesser;
        return result;
    }
};