find-most-frequent-vowel-and-consonant

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Algorithms

You are given a string <code>s</code> consisting of lowercase English letters (<code>&#39;a&#39;</code> to <code>&#39;z&#39;</code>). 

Your task is to:

<ul>
	<li>Find the vowel (one of <code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, or <code>&#39;u&#39;</code>) with the maximum frequency.</li>
	<li>Find the consonant (all other letters excluding vowels) with the maximum frequency.</li>
</ul>

Return the sum of the two frequencies.

Note: If multiple vowels or consonants have the same maximum frequency, you may choose any one of them. If there are no vowels or no consonants in the string, consider their frequency as 0.
The frequency of a letter <code>x</code> is the number of times it occurs in the string.
&nbsp;
Example 1:

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

Output: 6

Explanation:

<ul>
	<li>The vowels are: <code>&#39;u&#39;</code> (frequency 1), <code>&#39;e&#39;</code> (frequency 2). The maximum frequency is 2.</li>
	<li>The consonants are: <code>&#39;s&#39;</code> (frequency 4), <code>&#39;c&#39;</code> (frequency 2). The maximum frequency is 4.</li>
	<li>The output is <code>2 + 4 = 6</code>.</li>
</ul>
</div>

Example 2:

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

Output: 3

Explanation:

<ul>
	<li>The vowels are: <code>&#39;a&#39;</code> (frequency 3), <code>&#39;e&#39;</code> ( frequency 2), <code>&#39;i&#39;</code> (frequency 2). The maximum frequency is 3.</li>
	<li>There are no consonants in <code>s</code>. Hence, maximum consonant frequency = 0.</li>
	<li>The output is <code>3 + 0 = 3</code>.</li>
</ul>
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

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maxFreqSum

中级工程师

Find Most Frequent Vowel and Consonant

Hash Table

String

Counting

给你一个由小写英文字母（<code>'a'</code> 到 <code>'z'</code>）组成的字符串 <code>s</code>。你的任务是找出出现频率&nbsp;最高&nbsp;的元音（<code>'a'</code>、<code>'e'</code>、<code>'i'</code>、<code>'o'</code>、<code>'u'</code> 中的一个）和出现频率最高的辅音（除元音以外的所有字母），并返回这两个频数之和。

注意：如果有多个元音或辅音具有相同的最高频率，可以任选其中一个。如果字符串中没有元音或没有辅音，则其频率视为 0。
一个字母 <code>x</code> 的&nbsp;频率&nbsp;是它在字符串中出现的次数。

&nbsp;

示例 1：

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

输出: 6

解释:

<ul>
	<li>元音有：<code>'u'</code> 出现 1 次，<code>'e'</code> 出现 2 次。最大元音频率 = 2。</li>
	<li>辅音有：<code>'s'</code> 出现 4 次，<code>'c'</code> 出现 2 次。最大辅音频率 = 4。</li>
	<li>输出为 <code>2 + 4 = 6</code>。</li>
</ul>
</div>

示例 2：

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

输出: 3

解释:

<ul>
	<li>元音有：<code>'a'</code> 出现 3 次，<code>'e'</code> 出现 2 次，<code>'i'</code> 出现 2 次。最大元音频率 = 3。</li>
	<li><code>s</code> 中没有辅音。因此，最大辅音频率 = 0。</li>
	<li>输出为 <code>3 + 0 = 3</code>。</li>
</ul>
</div>

&nbsp;

提示:

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> 只包含小写英文字母</li>
</ul>