find-the-k-th-character-in-string-game-i

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Algorithms

Alice and Bob are playing a game. Initially, Alice has a string <code>word = &quot;a&quot;</code>.

You are given a positive integer <code>k</code>.

Now Bob will ask Alice to perform the following operation forever:

<ul>
	<li>Generate a new string by changing each character in <code>word</code> to its next character in the English alphabet, and append it to the original <code>word</code>.</li>
</ul>

For example, performing the operation on <code>&quot;c&quot;</code> generates <code>&quot;cd&quot;</code> and performing the operation on <code>&quot;zb&quot;</code> generates <code>&quot;zbac&quot;</code>.

Return the value of the <code>kth</code> character in <code>word</code>, after enough operations have been done for <code>word</code> to have at least <code>k</code> characters.

Note that the character <code>&#39;z&#39;</code> can be changed to <code>&#39;a&#39;</code> in the operation.

&nbsp;
Example 1:

<div class="example-block">
Input: k = 5

Output: &quot;b&quot;

Explanation:

Initially, <code>word = &quot;a&quot;</code>. We need to do the operation three times:

<ul>
	<li>Generated string is <code>&quot;b&quot;</code>, <code>word</code> becomes <code>&quot;ab&quot;</code>.</li>
	<li>Generated string is <code>&quot;bc&quot;</code>, <code>word</code> becomes <code>&quot;abbc&quot;</code>.</li>
	<li>Generated string is <code>&quot;bccd&quot;</code>, <code>word</code> becomes <code>&quot;abbcbccd&quot;</code>.</li>
</ul>
</div>

Example 2:

<div class="example-block">
Input: k = 10

Output: &quot;c&quot;
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= k &lt;= 500</code></li>
</ul>

kthCharacter

Find the K-th Character in String Game I

Shifting Letters

shifting-letters

字母移位

Bit Manipulation

Recursion

Math

Simulation

Alice 和 Bob 正在玩一个游戏。最初，Alice 有一个字符串 <code>word = "a"</code>。

给定一个正整数 <code>k</code>。

现在 Bob 会要求 Alice 执行以下操作 无限次 :

<ul>
	<li>将 <code>word</code> 中的每个字符 更改 为英文字母表中的 下一个 字符来生成一个新字符串，并将其 追加 到原始的 <code>word</code>。</li>
</ul>

例如，对 <code>"c"</code> 进行操作生成 <code>"cd"</code>，对 <code>"zb"</code> 进行操作生成 <code>"zbac"</code>。

在执行足够多的操作后， <code>word</code> 中 至少 存在 <code>k</code> 个字符，此时返回 <code>word</code> 中第 <code>k</code> 个字符的值。

注意，在操作中字符 <code>'z'</code> 可以变成 <code>'a'</code>。

&nbsp;

示例 1:

<div class="example-block">
输入：k = 5

输出："b"

解释：

最初，<code>word = "a"</code>。需要进行三次操作:

<ul>
	<li>生成的字符串是 <code>"b"</code>，<code>word</code> 变为 <code>"ab"</code>。</li>
	<li>生成的字符串是 <code>"bc"</code>，<code>word</code> 变为 <code>"abbc"</code>。</li>
	<li>生成的字符串是 <code>"bccd"</code>，<code>word</code> 变为 <code>"abbcbccd"</code>。</li>
</ul>
</div>

示例 2:

<div class="example-block">
输入：k = 10

输出："c"
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= k &lt;= 500</code></li>
</ul>