<p>Two strings are considered <strong>close</strong> if you can attain one from the other using the following operations:</p>

<ul>
	<li>Operation 1: Swap any two <strong>existing</strong> characters.

	<ul>
		<li>For example, <code>a<u>b</u>cd<u>e</u> -&gt; a<u>e</u>cd<u>b</u></code></li>
	</ul>
	</li>
	<li>Operation 2: Transform <strong>every</strong> occurrence of one <strong>existing</strong> character into another <strong>existing</strong> character, and do the same with the other character.
	<ul>
		<li>For example, <code><u>aa</u>c<u>abb</u> -&gt; <u>bb</u>c<u>baa</u></code> (all <code>a</code>&#39;s turn into <code>b</code>&#39;s, and all <code>b</code>&#39;s turn into <code>a</code>&#39;s)</li>
	</ul>
	</li>
</ul>

<p>You can use the operations on either string as many times as necessary.</p>

<p>Given two strings, <code>word1</code> and <code>word2</code>, return <code>true</code><em> if </em><code>word1</code><em> and </em><code>word2</code><em> are <strong>close</strong>, and </em><code>false</code><em> otherwise.</em></p>

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

<pre>
<strong>Input:</strong> word1 = &quot;abc&quot;, word2 = &quot;bca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can attain word2 from word1 in 2 operations.
Apply Operation 1: &quot;a<u>bc</u>&quot; -&gt; &quot;a<u>cb</u>&quot;
Apply Operation 1: &quot;<u>a</u>c<u>b</u>&quot; -&gt; &quot;<u>b</u>c<u>a</u>&quot;
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;a&quot;, word2 = &quot;aa&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>It is impossible to attain word2 from word1, or vice versa, in any number of operations.
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;cabbba&quot;, word2 = &quot;abbccc&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can attain word2 from word1 in 3 operations.
Apply Operation 1: &quot;ca<u>b</u>bb<u>a</u>&quot; -&gt; &quot;ca<u>a</u>bb<u>b</u>&quot;
Apply Operation 2: &quot;<u>c</u>aa<u>bbb</u>&quot; -&gt; &quot;<u>b</u>aa<u>ccc</u>&quot;
Apply Operation 2: &quot;<u>baa</u>ccc&quot; -&gt; &quot;<u>abb</u>ccc&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>word1</code> and <code>word2</code> contain only lowercase English letters.</li>
</ul>

class Solution {
public:
    bool closeStrings(string word1, string word2) {
        if(word1.size()==word2.size()){
            int mp1[256]={0};
            int mp2[256]={0};

            set<char>st1;
            set<char>st2;

            for(int i=0;i<word1.size();i++){
                mp1[word1[i]]++;
                st1.insert(word1[i]);
            }

            for(int i=0;i<word2.size();i++){
                mp2[word2[i]]++;
                st2.insert(word2[i]);
            }
            
            if(st1!=st2){
                return false;
            }

            sort(mp1,mp1+256,greater<int>());
            sort(mp2,mp2+256,greater<int>());

            return equal(begin(mp1), end(mp1), begin(mp2));

            } 

            return false;    
    }
};
23100