<p>Given two strings <code>s</code> and <code>goal</code>, return <code>true</code><em> if you can swap two letters in </em><code>s</code><em> so the result is equal to </em><code>goal</code><em>, otherwise, return </em><code>false</code><em>.</em></p>

<p>Swapping letters is defined as taking two indices <code>i</code> and <code>j</code> (0-indexed) such that <code>i != j</code> and swapping the characters at <code>s[i]</code> and <code>s[j]</code>.</p>

<ul>
	<li>For example, swapping at indices <code>0</code> and <code>2</code> in <code>&quot;abcd&quot;</code> results in <code>&quot;cbad&quot;</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> s = &quot;ab&quot;, goal = &quot;ba&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can swap s[0] = &#39;a&#39; and s[1] = &#39;b&#39; to get &quot;ba&quot;, which is equal to goal.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ab&quot;, goal = &quot;ab&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> The only letters you can swap are s[0] = &#39;a&#39; and s[1] = &#39;b&#39;, which results in &quot;ba&quot; != goal.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, goal = &quot;aa&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can swap s[0] = &#39;a&#39; and s[1] = &#39;a&#39; to get &quot;aa&quot;, which is equal to goal.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length, goal.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>s</code> and <code>goal</code> consist of lowercase letters.</li>
</ul>


class Solution {
public:
    bool buddyStrings(string s, string p) {
        int n=s.length(), m=p.length();
        if(n!=m) return false;
        
        vector<int> freq(26,0);
        
        int first=-1, second=-1;

        for(int i=0;i<n;i++){
            if(s[i] != p[i]){
                if(first==-1){
                    first=i;
                }else if(second==-1){
                    second=i;
                }else{
                    return false; // more than 2 mismatching characters
                }
            }
            freq[s[i]-'a']++; 
        }
        
        if(first == -1 && second == -1){
            for(int i=0;i<26;i++) if(freq[i] > 1) return true; // can swap similar characters
            return false; 
        }else if(second == -1){
            return false; // exactly 1 mismatch
        }
        
        return (s[first]==p[second] && s[second]==p[first]); // check similarity on 2 mismatched indexes
    }
};

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