<p>A <strong>sentence</strong> is a string of single-space separated words where each word consists only of lowercase letters.</p>

<p>A word is <strong>uncommon</strong> if it appears exactly once in one of the sentences, and <strong>does not appear</strong> in the other sentence.</p>

<p>Given two <strong>sentences</strong> <code>s1</code> and <code>s2</code>, return <em>a list of all the <strong>uncommon words</strong></em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s1 = &quot;this apple is sweet&quot;, s2 = &quot;this apple is sour&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">[&quot;sweet&quot;,&quot;sour&quot;]</span></p>

<p><strong>Explanation:</strong></p>

<p>The word <code>&quot;sweet&quot;</code> appears only in <code>s1</code>, while the word <code>&quot;sour&quot;</code> appears only in <code>s2</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s1 = &quot;apple apple&quot;, s2 = &quot;banana&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">[&quot;banana&quot;]</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 200</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters and spaces.</li>
	<li><code>s1</code> and <code>s2</code> do not have leading or trailing spaces.</li>
	<li>All the words in <code>s1</code> and <code>s2</code> are separated by a single space.</li>
</ul>

class Solution:
    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:
        c = s1.split() + s2.split()
        b = []
        for i in c:
            if c.count(i) == 1:
                b.append(i)
        return b
        # a = s1.split()
        # b = s2.split()
        # new_array1 = []
        # new_array2 = []
        # new_array3 = []
        # for i in a:
        #     if i not in new_array1:
        #         new_array1.append(i)
        # for j in b:
        #     if j not in new_array2:
        #         new_array2.append(j)
        # minimum = min(len(new_array1), len(new_array2))
        # i = 0
        # if len(a) == len(new_array1) and len(b) == len(new_array2):
        #     for i in range(minimum):
        #         if new_array1[i] != new_array2[i]:
        #             new_array3.append(new_array1[i])
        #             new_array3.append(new_array2[i])
        #     return new_array3
        # elif len(a) != len(new_array1) or len(b) != len(new_array2):
        #     for i in range(minimum):
        #         if new_array1[i] != new_array2[i]:
        #             new_array3.append(new_array2[i])
        #     return new_array3    
16500