- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, String, Dynamic Programming, leetcode-2606, Greedy, O(n), O(1), Kadane's Algorithm
Problem
You are given a string s
, a string chars
of distinct characters and an integer array vals
of the same length as chars
.
The cost of the substring is the sum of the values of each character in the substring. The cost of an empty string is considered 0
.
The value of the character is defined in the following way:
- If the character is not in the string
chars
, then its value is its corresponding position (1-indexed) in the alphabet.- For example, the value of
'a'
is1
, the value of'b'
is2
, and so on. The value of'z'
is26
.
- For example, the value of
- Otherwise, assuming
i
is the index where the character occurs in the stringchars
, then its value isvals[i]
.
Return the maximum cost among all substrings of the string s
.
Example 1:
Input: s = "adaa", chars = "d", vals = [-1000] Output: 2 Explanation: The value of the characters "a" and "d" is 1 and -1000 respectively. The substring with the maximum cost is "aa" and its cost is 1 + 1 = 2. It can be proven that 2 is the maximum cost.
Example 2:
Input: s = "abc", chars = "abc", vals = [-1,-1,-1] Output: 0 Explanation: The value of the characters "a", "b" and "c" is -1, -1, and -1 respectively. The substring with the maximum cost is the empty substring "" and its cost is 0. It can be proven that 0 is the maximum cost.
Constraints:
1 <= s.length <= 105
s
consist of lowercase English letters.1 <= chars.length <= 26
chars
consist of distinct lowercase English letters.vals.length == chars.length
-1000 <= vals[i] <= 1000