<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 goal) {
        if(s.size()!=goal.size()) return 0;
        int n = s.size();
        vector<int> v1(n, 0), v2(26, 0), v3(26, 0);

        for(int i = 0; i<n; i++){
            v2[s[i]-'a']++;
            v3[goal[i]-'a']++;
            if(s[i]==goal[i]) v1[i] = 1;
        }

        if(s==goal){
            for(int i = 0; i<26; i++){
                if(v2[i]>=2) return 1;
            }
            return 0;
        }

        int count = 0; //mismatch
        for(int i = 0; i<n; i++){
            if(!v1[i]) count++;
            if(count>2) return 0;
        }

        for(int i = 0; i<26; i++){
            if(v2[i]!=v3[i]) return 0;
        }

        return 1;
    }
};
8800