<p>You are given two strings <code>a</code> and <code>b</code> that consist of lowercase letters. In one operation, you can change any character in <code>a</code> or <code>b</code> to <strong>any lowercase letter</strong>.</p>

<p>Your goal is to satisfy <strong>one</strong> of the following three conditions:</p>

<ul>
	<li><strong>Every</strong> letter in <code>a</code> is <strong>strictly less</strong> than <strong>every</strong> letter in <code>b</code> in the alphabet.</li>
	<li><strong>Every</strong> letter in <code>b</code> is <strong>strictly less</strong> than <strong>every</strong> letter in <code>a</code> in the alphabet.</li>
	<li><strong>Both</strong> <code>a</code> and <code>b</code> consist of <strong>only one</strong> distinct letter.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> number of operations needed to achieve your goal.</em></p>

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

<pre>
<strong>Input:</strong> a = &quot;aba&quot;, b = &quot;caa&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Consider the best way to make each condition true:
1) Change b to &quot;ccc&quot; in 2 operations, then every letter in a is less than every letter in b.
2) Change a to &quot;bbb&quot; and b to &quot;aaa&quot; in 3 operations, then every letter in b is less than every letter in a.
3) Change a to &quot;aaa&quot; and b to &quot;aaa&quot; in 2 operations, then a and b consist of one distinct letter.
The best way was done in 2 operations (either condition 1 or condition 3).
</pre>

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

<pre>
<strong>Input:</strong> a = &quot;dabadd&quot;, b = &quot;cda&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The best way is to make condition 1 true by changing b to &quot;eee&quot;.
</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</code> and <code>b</code> consist only of lowercase letters.</li>
</ul>

class Solution 
{
public:
    int minCharacters(string a, string b) 
    {
        int ans=a.size()+b.size();
        for(char ch='a'; ch<'z'; ch++)
        {
            int op=0;
            for(int x=0; x<a.size(); x++)
            {
                if(a[x]>ch) op++;
            }
            for(int x=0; x<b.size(); x++)
            {
                if(ch>=b[x]) op++;
            }
            ans=min(ans,op);
        }
        for(char ch='a'; ch<'z'; ch++)
        {
            int op=0;
            for(int x=0; x<b.size(); x++)
            {
                if(b[x]>ch) op++;
            }
            for(int x=0; x<a.size(); x++)
            {
                if(ch>=a[x]) op++;
            }
            ans=min(ans,op);
        }
        map<char,int> am, bm;
        for(auto ch: a) am[ch]++;
        for(auto ch: b) bm[ch]++;
        int mxa=0;
        char cha='-';
        for(auto it: am) 
        {
            if(mxa<it.second)
            {
                mxa=it.second;
                cha=it.first;
            }
        }
        int mxb=0;
        char chb='-';
        for(auto it: bm) 
        {
            if(mxb<it.second)
            {
                mxb=it.second;
                chb=it.first;
            }
        }
        int aa=b.size()-mxb+(int)a.size()-am[chb];
        int bb=a.size()-mxa+(int)b.size()-bm[cha];
        ans=min(ans,min(aa,bb));
        return ans;
    }
};
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