<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 {
public:
    bool validPalindrome(string s) {
        int tot{};

        int i{}, j = s.size() - 1;
        while (i <= j) {
            if (s[i] != s[j]) {
                // cout << s[i] << " is not " << s[j] << endl;
                ++tot;
                if (isPalindrome(s, i, j - 1))
                    --j;

                else if (isPalindrome(s, i + 1, j))
                    ++i;

                else
                    return false;
            }
            ++i;
            --j;
        }

        return tot <= 1;
    }

    bool isPalindrome(string s, int start, int end) {
        int i = start, j = end;

        while (i <= j) {
            // if (!isalnum(s[i])) {++i; continue;}
            // if (!isalnum(s[j])) {--j; continue;}
            if (s[i] != s[j])
                return false;
            ++i;
            --j;
        }
        return true;
    }
};
23773