<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;
                }
            }
            freq[s[i] - 'a']++; 
        }
        
        if (first == -1 && second == -1) {
            for (int i = 0; i < 26; i++) if (freq[i] > 1) return true;
            return false; 
        } else if (second == -1) {
            return false;
        }
        
        return (s[first] == p[second] && s[second] == p[first]);
    }
};

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