<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 strSimilar(string s, string t) {
        reverse(t.begin(), t.end());
        int flag = 0;
        int i = 0, j = 0;
        int len = s.length();
        while(i < len) {
            if(s[i] != t[j]) {
                if(flag == 0) {
                    flag = 1;
                    j++;
                    continue;
                }
                else {
                    return 0;
                }
            }
            i++, j++; 
        }
        return 1;
    }
    bool validPalindrome(string s) {
        int len = s.length();
        if(len % 2 == 0) {
            int mid = len / 2;
            return strSimilar(s.substr(mid + 1), s.substr(0, mid)) || strSimilar(s.substr(0, mid - 1), s.substr(mid));
        }
        else {
            int mid = len / 2;
            return strSimilar(s.substr(mid + 1), s.substr(0, mid + 1)) || strSimilar(s.substr(0, mid), s.substr(mid));
        }
    }
};
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