{ "data": { "question": { "questionId": "3634", "questionFrontendId": "3412", "categoryTitle": "Algorithms", "boundTopicId": 3037803, "title": "Find Mirror Score of a String", "titleSlug": "find-mirror-score-of-a-string", "content": "
You are given a string s.
We define the mirror of a letter in the English alphabet as its corresponding letter when the alphabet is reversed. For example, the mirror of 'a' is 'z', and the mirror of 'y' is 'b'.
Initially, all characters in the string s are unmarked.
You start with a score of 0, and you perform the following process on the string s:
i, find the closest unmarked index j such that j < i and s[j] is the mirror of s[i]. Then, mark both indices i and j, and add the value i - j to the total score.j exists for the index i, move on to the next index without making any changes.Return the total score at the end of the process.
\n\n\n
Example 1:
\n\nInput: s = "aczzx"
\n\nOutput: 5
\n\nExplanation:
\n\ni = 0. There is no index j that satisfies the conditions, so we skip.i = 1. There is no index j that satisfies the conditions, so we skip.i = 2. The closest index j that satisfies the conditions is j = 0, so we mark both indices 0 and 2, and then add 2 - 0 = 2 to the score.i = 3. There is no index j that satisfies the conditions, so we skip.i = 4. The closest index j that satisfies the conditions is j = 1, so we mark both indices 1 and 4, and then add 4 - 1 = 3 to the score.Example 2:
\n\nInput: s = "abcdef"
\n\nOutput: 0
\n\nExplanation:
\n\nFor each index i, there is no index j that satisfies the conditions.
\n
Constraints:
\n\n1 <= s.length <= 105s consists only of lowercase English letters.给你一个字符串 s。
英文字母中每个字母的 镜像 定义为反转字母表之后对应位置上的字母。例如,'a' 的镜像是 'z','y' 的镜像是 'b'。
最初,字符串 s 中的所有字符都 未标记 。
字符串 s 的初始分数为 0 ,你需要对其执行以下过程:
i ,找到距离最近的 未标记 下标 j,下标 j 需要满足 j < i 且 s[j] 是 s[i] 的镜像。然后 标记 下标 i 和 j,总分加上 i - j 的值。i,不存在满足条件的下标 j,则跳过该下标,继续处理下一个下标,不需要进行标记。返回最终的总分。
\n\n\n\n
示例 1:
\n\n输入: s = \"aczzx\"
\n\n输出: 5
\n\n解释:
\n\ni = 0。没有符合条件的下标 j,跳过。i = 1。没有符合条件的下标 j,跳过。i = 2。距离最近的符合条件的下标是 j = 0,因此标记下标 0 和 2,然后将总分加上 2 - 0 = 2 。i = 3。没有符合条件的下标 j,跳过。i = 4。距离最近的符合条件的下标是 j = 1,因此标记下标 1 和 4,然后将总分加上 4 - 1 = 3 。示例 2:
\n\n输入: s = \"abcdef\"
\n\n输出: 0
\n\n解释:
\n\n对于每个下标 i,都不存在满足条件的下标 j。
\n\n
提示:
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