<p>You are given two strings <code>a</code> and <code>b</code> of the same length. Choose an index and split both strings <strong>at the same index</strong>, splitting <code>a</code> into two strings: <code>a<sub>prefix</sub></code> and <code>a<sub>suffix</sub></code> where <code>a = a<sub>prefix</sub> + a<sub>suffix</sub></code>, and splitting <code>b</code> into two strings: <code>b<sub>prefix</sub></code> and <code>b<sub>suffix</sub></code> where <code>b = b<sub>prefix</sub> + b<sub>suffix</sub></code>. Check if <code>a<sub>prefix</sub> + b<sub>suffix</sub></code> or <code>b<sub>prefix</sub> + a<sub>suffix</sub></code> forms a palindrome.</p>

<p>When you split a string <code>s</code> into <code>s<sub>prefix</sub></code> and <code>s<sub>suffix</sub></code>, either <code>s<sub>suffix</sub></code> or <code>s<sub>prefix</sub></code> is allowed to be empty. For example, if <code>s = &quot;abc&quot;</code>, then <code>&quot;&quot; + &quot;abc&quot;</code>, <code>&quot;a&quot; + &quot;bc&quot;</code>, <code>&quot;ab&quot; + &quot;c&quot;</code> , and <code>&quot;abc&quot; + &quot;&quot;</code> are valid splits.</p>

<p>Return <code>true</code><em> if it is possible to form</em><em> a palindrome string, otherwise return </em><code>false</code>.</p>

<p><strong>Notice</strong> that&nbsp;<code>x + y</code> denotes the concatenation of strings <code>x</code> and <code>y</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> a = &quot;x&quot;, b = &quot;y&quot;
<strong>Output:</strong> true
<strong>Explaination:</strong> If either a or b are palindromes the answer is true since you can split in the following way:
a<sub>prefix</sub> = &quot;&quot;, a<sub>suffix</sub> = &quot;x&quot;
b<sub>prefix</sub> = &quot;&quot;, b<sub>suffix</sub> = &quot;y&quot;
Then, a<sub>prefix</sub> + b<sub>suffix</sub> = &quot;&quot; + &quot;y&quot; = &quot;y&quot;, which is a palindrome.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> a = &quot;xbdef&quot;, b = &quot;xecab&quot;
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> a = &quot;ulacfd&quot;, b = &quot;jizalu&quot;
<strong>Output:</strong> true
<strong>Explaination:</strong> Split them at index 3:
a<sub>prefix</sub> = &quot;ula&quot;, a<sub>suffix</sub> = &quot;cfd&quot;
b<sub>prefix</sub> = &quot;jiz&quot;, b<sub>suffix</sub> = &quot;alu&quot;
Then, a<sub>prefix</sub> + b<sub>suffix</sub> = &quot;ula&quot; + &quot;alu&quot; = &quot;ulaalu&quot;, which is a palindrome.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>5</sup></code></li>
	<li><code>a.length == b.length</code></li>
	<li><code>a</code> and <code>b</code> consist of lowercase English letters</li>
</ul>

class Solution 
{
    bool is_palindrome(string s)
    {
        for(int i{}; i<size(s)/2; ++i)
            if(s[i]!=s[size(s)-1-i]) return false;
        return true;
    }
public:
    bool checkPalindromeFormation(string a, string b) 
    {
        int i{}, j(size(b)-1);
        for(; i<j and a[i]==b[j]; ++i, --j);
        if(is_palindrome(string(begin(b)+i, begin(b)+j+1))) return true;
        
        i = 0; j = size(b)-1;
        for(; i<j and b[i]==a[j]; ++i, --j);
        if(is_palindrome(string(begin(a)+i, begin(a)+j+1))) return true;
        
        int jj = size(b)-1, ii = 0;
        for(; jj>ii and a[jj]==b[ii]; --jj, ++ii);
        if(is_palindrome(string(begin(b)+ii, begin(b)+jj+1))) return true;
        
        i = size(b)-1; j = 0;
        for(; i>j and b[i]==a[j]; --i, ++j);
        if(is_palindrome(string(begin(a)+j, begin(a)+i+1))) return true;       
        return false;
    }
};




    
    
/*

 0123456 7
 a f  f d
 ||||||| *                           |||||||    
 u c  g a
                                   7 6543210 

    
         |
    a f f|d
         | 
    u c x|a
         |
         
             
             
             
*/              
    
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