<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:
        p1 = 0
        p2 = len(s) - 1
        self.s = s
        s += '*'
        print(self.s, s)
        while p1 < p2:
            if s[p1] == s[p2]:
                p1 += 1
                p2 -= 1
            else:
                return self.oneTrial(p1+1, p2) or self.oneTrial(p1, p2-1)
        return True

    def oneTrial(self, i, j):
        p1 = i
        p2 = j
        while p1 < p2:
            if self.s[p1] == self.s[p2]:
                p1 += 1
                p2 -= 1
            else:
                return False
        return True
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