<p>Given an array of strings <code>words</code>, return <em>the words that can be typed using letters of the alphabet on only one row of American keyboard like the image below</em>.</p>

<p>In the <strong>American keyboard</strong>:</p>

<ul>
	<li>the first row consists of the characters <code>&quot;qwertyuiop&quot;</code>,</li>
	<li>the second row consists of the characters <code>&quot;asdfghjkl&quot;</code>, and</li>
	<li>the third row consists of the characters <code>&quot;zxcvbnm&quot;</code>.</li>
</ul>
<img alt="" src="https://assets.leetcode.com/uploads/2018/10/12/keyboard.png" style="width: 800px; max-width: 600px; height: 267px;" />
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;Hello&quot;,&quot;Alaska&quot;,&quot;Dad&quot;,&quot;Peace&quot;]
<strong>Output:</strong> [&quot;Alaska&quot;,&quot;Dad&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;omk&quot;]
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;adsdf&quot;,&quot;sfd&quot;]
<strong>Output:</strong> [&quot;adsdf&quot;,&quot;sfd&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 20</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
	<li><code>words[i]</code> consists of English letters (both lowercase and uppercase).&nbsp;</li>
</ul>

class Solution:
    def findWords(self, words):
        m = words
        result = []
        a = ['q','w','e','r','t','y','u','i','o','p','Q','W','E','R','T','Y','U','I','O','P']
        b = ['a','s','d','f','g','h','j','k','l','A','S','D','F','G','H','J','K','L']
        c = ['z','x','c','v','b','n','m','Z','X','C','V','B','N','M']
        for i in m:
            i = list(i)
            ct1 = ct2 = ct3 = 0
            for j in range(len(i)):
                if i[j] in a:
                    ct1+=1
                elif i[j] in b:
                    ct2+=1
                elif i[j] in c:
                    ct3+=1
            if ct1==len(i) or ct2==len(i) or ct3==len(i):
                i = ''.join(i)
                result.append(i)
        return result
16400