<p>Given a string <code>s</code>, return <code>true</code> <em>if the </em><code>s</code><em> can be palindrome after deleting <strong>at most one</strong> character from it</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> true
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You could delete the character &#39;c&#39;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def validPalindrome(self, s: str) -> bool:
        arr1 = [x for x in s]
        arr2 = []
        arr3 = []

        firstMistake = False

        for i in range(len(arr1) // 2):
            if arr1[i] != arr1[- i - 1]:
                firstMistake = True
                for j in range(len(arr1)):
                    if j != i:
                        arr2.append(arr1[j])

                for j in range(len(arr1)):
                    if j != len(arr1) - i - 1:
                        arr3.append(arr1[j])

                break

        errors = 0
        if firstMistake:
            for i in range(len(arr2) // 2):
                if arr2[i] != arr2[- i - 1]:
                    errors += 1
                    break

            for i in range(len(arr3) // 2):
                if arr3[i] != arr3[- i - 1]:
                    errors += 1
                    break

        if errors == 2:
            return False

        return True
        
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