<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 solve(string a,string b)
    {
        vector<int> freqA(26,0);
        vector<int> freqB(26,0);

        for(auto c:a)
        {
            freqA[c-'a']++;
        }

        for(auto c:b)
        {
            freqB[c-'a']++;
        }
        int ans12 = a.length()+b.length();
        for(int i=1;i<26;i++)
        {
            int opern = 0;
            for(int j=i;j<26;j++)
            {
                opern+=freqA[j];
            }

            for(int j=0;j<i;j++)
            {
                opern+=freqB[j];
            }

            ans12 = min(ans12,opern);
        }

        return ans12;
    }
public:
    int minCharacters(string a, string b) {
        int a1 = solve(a,b);
        int a2 = solve(b,a);
        vector<int> freqA(26,0);
        vector<int> freqB(26,0);

        for(auto c:a)
        {
            freqA[c-'a']++;
        }

        for(auto c:b)
        {
            freqB[c-'a']++;
        }
        int m = a.length()+b.length();
        int ans = a.length()+b.length();
        for(int i=0;i<26;i++)
        {
            ans = min(ans,m-freqA[i]-freqB[i]);
        }

        return min({ans,a1,a2});
    }
};
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