find-mirror-score-of-a-string

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Algorithms

You are given a string <code>s</code>.

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 <code>&#39;a&#39;</code> is <code>&#39;z&#39;</code>, and the mirror of <code>&#39;y&#39;</code> is <code>&#39;b&#39;</code>.

Initially, all characters in the string <code>s</code> are unmarked.

You start with a score of 0, and you perform the following process on the string <code>s</code>:

<ul>
	<li>Iterate through the string from left to right.</li>
	<li>At each index <code>i</code>, find the closest unmarked index <code>j</code> such that <code>j &lt; i</code> and <code>s[j]</code> is the mirror of <code>s[i]</code>. Then, mark both indices <code>i</code> and <code>j</code>, and add the value <code>i - j</code> to the total score.</li>
	<li>If no such index <code>j</code> exists for the index <code>i</code>, move on to the next index without making any changes.</li>
</ul>

Return the total score at the end of the process.

&nbsp;
Example 1:

<div class="example-block">
Input: s = &quot;aczzx&quot;

Output: 5

Explanation:

<ul>
	<li><code>i = 0</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>
	<li><code>i = 1</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>
	<li><code>i = 2</code>. The closest index <code>j</code> that satisfies the conditions is <code>j = 0</code>, so we mark both indices 0 and 2, and then add <code>2 - 0 = 2</code> to the score.</li>
	<li><code>i = 3</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>
	<li><code>i = 4</code>. The closest index <code>j</code> that satisfies the conditions is <code>j = 1</code>, so we mark both indices 1 and 4, and then add <code>4 - 1 = 3</code> to the score.</li>
</ul>
</div>

Example 2:

<div class="example-block">
Input: s = &quot;abcdef&quot;

Output: 0

Explanation:

For each index <code>i</code>, there is no index <code>j</code> that satisfies the conditions.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 105</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

calculateScore

Find Mirror Score of a String

Stack

Hash Table

String

Simulation

给你一个字符串 <code>s</code>。

英文字母中每个字母的&nbsp;镜像&nbsp;定义为反转字母表之后对应位置上的字母。例如，<code>'a'</code> 的镜像是 <code>'z'</code>，<code>'y'</code> 的镜像是 <code>'b'</code>。

最初，字符串 <code>s</code> 中的所有字符都&nbsp;未标记&nbsp;。

字符串 <code>s</code>&nbsp;的初始分数为 0 ，你需要对其执行以下过程：

<ul>
	<li>从左到右遍历字符串。</li>
	<li>对于每个下标&nbsp;<code>i&nbsp;</code>，找到距离最近的&nbsp;未标记 下标&nbsp;<code>j</code>，下标 <code>j</code> 需要满足&nbsp;<code>j &lt; i</code> 且 <code>s[j]</code> 是 <code>s[i]</code> 的镜像。然后&nbsp;标记 下标&nbsp;<code>i</code> 和 <code>j</code>，总分加上&nbsp;<code>i - j</code>&nbsp;的值。</li>
	<li>如果对于下标&nbsp;<code>i</code>，不存在满足条件的下标&nbsp;<code>j</code>，则跳过该下标，继续处理下一个下标，不需要进行标记。</li>
</ul>

返回最终的总分。

&nbsp;

示例 1：

<div class="example-block">
输入： s = "aczzx"

输出： 5

解释：

<ul>
	<li><code>i = 0</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>
	<li><code>i = 1</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>
	<li><code>i = 2</code>。距离最近的符合条件的下标是 <code>j = 0</code>，因此标记下标&nbsp;0 和 2，然后将总分加上&nbsp;<code>2 - 0 = 2</code>&nbsp;。</li>
	<li><code>i = 3</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>
	<li><code>i = 4</code>。距离最近的符合条件的下标是 <code>j = 1</code>，因此标记下标&nbsp;1 和 4，然后将总分加上&nbsp;<code>4 - 1 = 3</code>&nbsp;。</li>
</ul>
</div>

示例 2：

<div class="example-block">
输入： s = "abcdef"

输出： 0

解释：

对于每个下标&nbsp;<code>i</code>，都不存在满足条件的下标&nbsp;<code>j</code>。
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 105</code></li>
	<li><code>s</code> 仅由小写英文字母组成。</li>
</ul>