minimum-operations-to-make-a-rotated-palindrome-i

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Algorithms

You are given a string <code>s</code> consisting of lowercase English letters.

You can perform the following operations any number of times (including zero) and in any order:

<ul>
	<li>Increment: Choose any index <code>i</code> and replace <code>s[i]</code> with the next lowercase English letter. The letter after <code>&#39;z&#39;</code> is <code>&#39;a&#39;</code>.</li>
	<li>Left rotate: Move the first character of the string to the end.</li>
</ul>

Return the minimum number of operations required to make <code>s</code> a palindrome.

&nbsp;
Example 1:

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

Output: 2

Explanation:
One optimal solution:

<ul>
	<li>Left rotate the string: <code>&quot;abc&quot; -&gt; &quot;bca&quot;</code>.</li>
	<li>Increment <code>&#39;a&#39;</code> to <code>&#39;b&#39;</code>: <code>&quot;bca&quot; -&gt; &quot;bcb&quot;</code>.</li>
	<li><code>&quot;bcb&quot;</code> is a palindrome. Thus, the answer is 2.</li>
</ul>
</div>

Example 2:

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

Output: 3

Explanation:

<ul>
	<li>Increment the first character three times: <code>&quot;yb&quot; -&gt; &quot;zb&quot; -&gt; &quot;ab&quot; -&gt; &quot;bb&quot;</code>.</li>
	<li><code>&quot;bb&quot;</code> is a palindrome. Thus, the answer is 3.</li>
</ul>
</div>

&nbsp;
Constraints:

<ul>
	<li><code>2 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

Biweekly Contest 189

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Minimum Operations to Make a Rotated Palindrome I

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vanillajs

You are given a string s consisting of lowercase English letters.

You can perform the following operations any number of times (including zero) and in any order:

 * Increment: Choose any index i and replace s[i] with the next lowercase English letter. The letter after 'z' is 'a'.
 * Left rotate: Move the first character of the string to the end.

Return the minimum number of operations required to make s a palindrome.

 

Example 1:

Input: s = "abc"

Output: 2

Explanation:

One optimal solution:
 * Left rotate the string: "abc" -> "bca".
 * Increment 'a' to 'b': "bca" -> "bcb".
 * "bcb" is a palindrome. Thus, the answer is 2.

Example 2:

Input: s = "yb"

Output: 3

Explanation:

 * Increment the first character three times: "yb" -> "zb" -> "ab" -> "bb".
 * "bb" is a palindrome. Thus, the answer is 3.

 

Constraints:

 * 2 <= s.length <= 2000
 * s consists only of lowercase English letters.