Make Number of Distinct Characters Equal
<p>You are given two <strong>0-indexed</strong> strings <code>word1</code> and <code>word2</code>.</p>

<p>A <strong>move</strong> consists of choosing two indices <code>i</code> and <code>j</code> such that <code>0 &lt;= i &lt; word1.length</code> and <code>0 &lt;= j &lt; word2.length</code> and swapping <code>word1[i]</code> with <code>word2[j]</code>.</p>

<p>Return <code>true</code> <em>if it is possible to get the number of distinct characters in</em> <code>word1</code> <em>and</em> <code>word2</code> <em>to be equal with <strong>exactly one</strong> move. </em>Return <code>false</code> <em>otherwise</em>.</p>

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

<pre>
<strong>Input:</strong> word1 = &quot;ac&quot;, word2 = &quot;b&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> Any pair of swaps would yield two distinct characters in the first string, and one in the second string.
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;abcc&quot;, word2 = &quot;aab&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = &quot;abac&quot; and word2 = &quot;cab&quot;, which both have 3 distinct characters.
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;abcde&quot;, word2 = &quot;fghij&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Both resulting strings will have 5 distinct characters, regardless of which indices we swap.
</pre>

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

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>word1</code> and <code>word2</code> consist of only lowercase English letters.</li>
</ul>
Hash Table; String; Counting
class Solution:
    def isItPossible(self, word1: str, word2: str) -> bool:
        cnt1 = Counter(word1)
        cnt2 = Counter(word2)
        x, y = len(cnt1), len(cnt2)
        for c1, v1 in cnt1.items():
            for c2, v2 in cnt2.items():
                if c1 == c2:
                    if x == y:
                        return True
                else:
                    a = x - (v1 == 1) + (cnt1[c2] == 0)
                    b = y - (v2 == 1) + (cnt2[c1] == 0)
                    if a == b:
                        return True
        return False
