<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):

        L, R = 0, (len(s)-1)

        cL = 0
        cR = 0
        while L <= R:
            print(s[L],s[R])
            print(cL, cR)
            if s[L] == s[R]:
                L += 1
                R -= 1
            elif s[L] != s[R] and cL ==0 and cR == 0:
                if s[L+1] == s[R]:
                    L += 1
                    cL += 1
                elif s[L] == s[R-1]:
                    R -= 1
                    cR += 1
                else: return (False)

            elif s[L] != s[R] and cL == 1:
                    L -= 2
                    cL += 1
            elif s[L] != s[R] and cR == 1:
                    R += 2
                    cR += 1

            else:
                return (False)

        return (True)
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