Valid Palindrome II
<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>
Greedy; Two Pointers; String
/**
 * @param {string} s
 * @return {boolean}
 */
var validPalindrome = function (s) {
    const check = function (i, j) {
        for (; i < j; ++i, --j) {
            if (s[i] !== s[j]) {
                return false;
            }
        }
        return true;
    };
    for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
        if (s[i] !== s[j]) {
            return check(i + 1, j) || check(i, j - 1);
        }
    }
    return true;
};
